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For each of the following unbalanced equations, suppose that exactly \(1.00 \mathrm{g}\) of each reactant is taken. Determine which reactant is limiting, and calculate what mass of the product in boldface is expected (assuming that the limiting reactant is completely consumed). a. \(\mathrm{CS}_{2}(l)+\mathrm{O}_{2}(g) \rightarrow \mathbf{C} \mathbf{O}_{2}(g)+\mathrm{SO}_{2}(g)\) b. \(\mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{CN}_{2} \mathrm{H}_{4} \mathrm{O}(s)+\mathbf{H}_{2} \mathbf{O}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{MnO}_{2}(s) \rightarrow \mathrm{MnO}(s)+\mathbf{H}_{2} \mathbf{O}(g)\) d. \(\mathrm{I}_{2}(l)+\mathrm{Cl}_{2}(g) \rightarrow \mathbf{I C l}(g)\)

Short Answer

Expert verified
The mass of the expected products in each of the reactions is as follows: a. \(0.458 \ g \ CO_2\) b. \(0.408 \ g \ H_2O\) c. \(0.207 \ g \ H_2O\) d. \(1.28 \ g \ ICl\)

Step by step solution

01

Balance the equation

The balanced equation is \[CS_2(l) + 3O_2(g) \rightarrow CO_2(g) + 2SO_2(g)\]
02

Determine the limiting reactant

To find the limiting reactant, calculate the moles of both reactants and then compare their mole ratios: \[CS_2: \frac{1.00 \ g}{76.14 \ g/mol} = 0.0131 \ mol\] \[O_2: \frac{1.00 \ g}{32.00 \ g/mol} = 0.03125 \ mol\] Mole ratio of CSâ‚‚ and Oâ‚‚ should be 1:3, but here it is \(0.0131:0.03125\), which is about 1:2.4. So, the limiting reactant is \(O_2\) because we have less of it relative to the stoichiometry.
03

Calculate the mass of the product

As the limiting reactant is Oâ‚‚, calculate the mass of COâ‚‚ produced: \[\frac{0.03125\ mol \ O_2}{3\ mol\ O_2} \times 1\ mol\ CO_2 \times 44.01\ g\ CO_2/mol\ CO_2 = 0.458\ g\ CO_2\] b. \(\mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{CN}_{2}\mathrm{H}_{4} \mathrm{O}(s)+\mathbf{H}_{2} \mathbf{O}(g)\)
04

Balance the equation

The balanced equation is \[2NH_3(g) + CO_2(g) \rightarrow CN_2H_4O(s) + H_2O(g)\]
05

Determine the limiting reactant

Calculate the moles of both reactants and then compare their mole ratios: \[NH_3: \frac{1.00 \ g}{17.03 \ g/mol} = 0.0587 \ mol\] \[CO_2: \frac{1.00 \ g}{44.01 \ g/mol} = 0.0227 \ mol\] Mole ratio of NH₃ and CO₂ should be 2:1, but here it is \(0.0587:0.0227\), which is about 2.6:1. So, the limiting reactant is \(CO_2\) because we have less of it relative to the stoichiometry.
06

Calculate the mass of the product

As the limiting reactant is COâ‚‚, calculate the mass of Hâ‚‚O produced: \[\frac{0.0227\ mol \ CO_2}{1\ mol\ CO_2} \times 1\ mol\ H_2O \times 18.02\ g\ H_2O/mol\ H_2O = 0.408\ g\ H_2O\] c. \(\mathrm{H}_{2}(g)+\mathrm{MnO}_{2}(s) \rightarrow \mathrm{MnO}(s)+\mathbf{H}_{2} \mathbf{O}(g)\)
07

Balance the equation

The balanced equation is \[H_2(g) + MnO_2(s) \rightarrow MnO(s) + H_2O(g)\] (This equation is impossible to balance due to violating charge conservation, however, we will assume the equation is correct and proceed to the next steps.)
08

Determine the limiting reactant

Calculate the moles of both reactants and then compare their mole ratios: \[H_2: \frac{1.00 \ g}{2.02 \ g/mol} = 0.495 \ mol\] \[MnO_2: \frac{1.00 \ g}{86.94 \ g/mol} = 0.0115 \ mol\] Mole ratio of Hâ‚‚ and MnOâ‚‚ should be 1:1. Here, the mole ratio is much higher for Hâ‚‚; thus, the limiting reactant is \(MnO_2\).
09

Calculate the mass of the product

As the limiting reactant is MnOâ‚‚, calculate the mass of Hâ‚‚O produced: \[\frac{0.0115\ mol \ MnO_2}{1\ mol\ MnO_2} \times 1\ mol\ H_2O \times 18.02\ g\ H_2O/mol\ H_2O = 0.207\ g\ H_2O\] d. \(\mathrm{I}_{2}(l)+\mathrm{Cl}_{2}(g) \rightarrow \mathbf{I C l}(g)\)
10

Balance the equation

The balanced equation is \[I_2(l) + Cl_2(g) \rightarrow 2ICl(g)\]
11

Determine the limiting reactant

Calculate the moles of both reactants and then compare their mole ratios: \[I_2: \frac{1.00 \ g}{253.8 \ g/mol} = 0.00394 \ mol\] \[Cl_2: \frac{1.00 \ g}{70.90 \ g/mol} = 0.0141 \ mol\] Mole ratio of Iâ‚‚ and Clâ‚‚ should be 1:1. Here, the mole ratio is much lower for Iâ‚‚; thus, the limiting reactant is \(I_2\).
12

Calculate the mass of the product

As the limiting reactant is Iâ‚‚, calculate the mass of ICl produced: \[\frac{0.00394\ mol \ I_2}{1\ mol\ I_2} \times 2\ mol\ ICl \times 162.35\ g\ ICl/mol\ ICl = 1.28\ g\ ICl\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry in Chemical Reactions
Stoichiometry is the foundation of chemistry. It involves the calculation of reactants and products in chemical reactions. By understanding stoichiometry, we can predict how much of each substance participates in a reaction. The key is mole ratio, which is derived from the coefficients in a balanced chemical equation.

In any reaction, the numbers of atoms for each element need to be equal on both sides of the equation. This conservation allows us to determine how much product is made from given reactants. The use of mole ratios from balanced equations lets us convert between grams, moles, and even molecules. This is essential for understanding how chemical reactions work on a quantitative level. When problems arise, such as finding the limiting reactant, stoichiometry guides us to the correct solution.

Using stoichiometry, students can understand the relationship between reactants and products, and see why some substances run out before others. It’s a crucial skill in chemistry that applies to various fields like pharmaceuticals, energy, and environmental science.
Balancing Chemical Equations
Balancing chemical equations is a vital step before performing any calculation in stoichiometry. An unbalanced equation doesn't accurately represent the reaction, making it impossible to correctly evaluate the amount of reactants and products.

To balance an equation:
  • Count the number of each type of atom on both sides of the equation.
  • Use coefficients, which are numbers placed before compounds, to equalize the atom count on each side.
  • Do not change the subscripts within compounds, as that alters the substance itself.
Balancing ensures that mass and charge are conserved, reflecting the real-world behavior of chemical reactions.

For instance, if you start with a reaction like \[CS_2(l) + O_2(g) \rightarrow CO_2(g) + SO_2(g)\]you need to ensure the same number of carbon, sulfur, and oxygen atoms before and after the reaction. This process is crucial for determining which reactant might limit the production of products, as it informs the stoichiometric calculations that follow.
Mole Calculation Techniques
Mole calculations are a core part of working with chemical equations. These calculations help convert between mass, moles, and even the number of molecules.

Consider these steps:
  • Determine the molar mass of each compound or element, using the atomic mass from the periodic table.
  • Calculate moles from grams using the formula \( \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \).
  • Use these mole amounts to find out the limiting reactant by comparing the actual mole ratio to the stoichiometric ratio.
These techniques are vital because they translate a chemist’s understanding from grams to the chemical amounts actually participating in the reaction. Once you have calculated the number of moles, you can solve most problems regarding reaction conclusions and predict the amount of product formed.

In practice, accurately conducting mole calculations ensures precision in experiments and industrial chemical processes, making it a vital skill for any chemistry student.
Understanding Reaction Products
Reaction products are the substances formed as a result of a chemical reaction. To predict what products will form, you need to know the limiting reactant, which is the reactant that gets entirely used up first, limiting the production of any further product.

After determining the limiting reactant, you can calculate the amount of each product that will form using stoichiometry. The key is to use the balanced chemical equation and apply mole-to-mole conversions between reactants and products.
  • Identify the limiting reactant through mole calculations.
  • Use the mole ratio from the balanced equation to find the amount of product formed.
  • Convert moles of product back to grams if needed using the molar mass.
For example, in the reaction \[I_2(l) + Cl_2(g) \rightarrow 2ICl(g)\], knowing which reactant is limiting helps us calculate the exact grams of \(ICl\) produced.

Understanding these concepts ensures that chemists can control reactions and optimize the yields of products, which is fundamental in labs and industries where precise quantities are crucial.

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Most popular questions from this chapter

When elemental carbon is burned in the open atmosphere, with plenty of oxygen gas present, the product is carbon dioxide. $$\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)$$ However, when the amount of oxygen present during the burning of the carbon is restricted, carbon monoxide is more likely to result. $$2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}(g)$$What mass of each product is expected when a \(5.00-\mathrm{g}\) sample of pure carbon is burned under each of these conditions?

Although we usually think of substances as "burning" only in oxygen gas, the process of rapid oxidation to produce a flame may also take place in other strongly oxidizing gases. For example, when iron is heated and placed in pure chlorine gas, the iron "burns" according to the following (unbalanced) reaction: $$\mathrm{Fe}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{FeCl}_{3}(s)$$ How many milligrams of iron(III) chloride result when \(15.5 \mathrm{mg}\) of iron is reacted with an excess of chlorine gas?

Before going to lab, a student read in his lab manual that the percent yield for a difficult reaction to be studied was likely to be only \(40 . \%\) of the theoretical yield. The student's prelab stoichiometric calculations predict that the theoretical yield should be 12.5 g. What is the student's actual yield likely to be?

What does it mean to say a reactant is present "in excess" in a process? Can the limiting reactant be present in excess? Does the presence of an excess of a reactant affect the mass of products expected for a reaction?

If steel wool (iron) is heated until it glows and is placed in a bottle containing pure oxygen, the iron reacts spectacularly to produce iron(III) oxide. $$\mathrm{Fe}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)$$ If \(1.25 \mathrm{g}\) of iron is heated and placed in a bottle containing 0.0204 mol of oxygen gas, what mass of iron(III) oxide is produced?

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