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Chapter 9: Problem 16

Calcium cyanamide, \(\mathrm{CaCN}_{2}\), reacts with water to form calcium carbonate and ammonia. $$ \mathrm{CaCN}_{2}(s)+3 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{CaCO}_{3}(s)+2 \mathrm{NH}_{3}(g) $$ a. How many grams of \(\mathrm{H}_{2} \mathrm{O}\) are needed to react with \(75.0 \mathrm{~g}\) of \(\mathrm{CaCN}_{2} ?\) b. How many grams of \(\mathrm{NH}_{3}\) are produced from \(5.24 \mathrm{~g}\) of \(\mathrm{CaCN}_{2} ?\) c. How many grams of \(\mathrm{CaCO}_{3}\) form if \(155 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) reacts?

Short Answer

Expert verified
a. 50.6 g \ \mathrm{H}_2Ob. 2.22 g \ \mathrm{NH}_3c. 287 g \ \mathrm{CaCO}_3

Step by step solution

01

Molecular Weights Calculation

Calculate the molecular weights:1. \(\mathrm{CaCN}_2: 40.08 + 2(12.01 + 14.01) = 80.10 \, \mathrm{g/mol}\)2. \(\mathrm{H}_2O: 2(1.01) + 16.00 = 18.02 \, \mathrm{g/mol}\)3. \(\mathrm{NH}_3: 14.01 + 3(1.01) = 17.04 \, \mathrm{g/mol}\)4. \(\mathrm{CaCO}_3: 40.08 + 12.01 + 3(16.00) = 100.09 \, \mathrm{g/mol}\)
02

Write Balanced Equation and Ratios

The balanced chemical equation is: \( \mathrm{CaCN}_2(s)+3 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{CaCO}_3(s)+ 2 \mathrm{NH}_3(g) \)From the equation:1 mol \ \mathrm{CaCN}_2 reacts with 3 mol \ \mathrm{H}_2O 1 mol \ \mathrm{CaCN}_2 produces 2 mol \ \mathrm{NH}_3 1 mol \ \mathrm{CaCN}_2 produces 1 mol \ \mathrm{CaCO}_3
03

Step 3a: Calculate grams of \( \mathrm{H}_2 \mathrm{O} \) needed for 75.0 g \( \mathrm{CaCN}_2 \)

Start with converting grams of \( \mathrm{CaCN}_2 \) to moles: \( n = \dfrac{75.0 \, \mathrm{g}}{80.10 \, \mathrm{g/mol}} = 0.936 \mathrm{mol} \, \mathrm{CaCN}_2\)Next, use the mole ratio from the balanced equation (3 mol \ \mathrm{H}_2O / 1 mol \ \mathrm{CaCN}_2) to find moles of \ \mathrm{H}_2O \( \mathrm{moles} \ \ \mathrm{H}_2O = 0.936 \, \mathrm{mol} \ \ \mathrm{CaCN}_2 \times 3 \frac{\mathrm{mol} \ \ \mathrm{H}_2O}{\mathrm{mol} \ \ \mathrm{CaCN}_2}= 2.81 \, \mathrm{mol} \ \ \mathrm{H}_2O\)Finally, convert moles of \ \mathrm{H}_2O to grams:\( \mathrm{grams} \ \ \mathrm{H}_2O = 2.81 \, \mathrm{mol} \times 18.02 \, \mathrm{g/mol} = 50.6 \, \mathrm{g} \ \ \mathrm{H}_2O\)
04

Step 3b: Calculate grams of \( \mathrm{NH}_3 \) produced from 5.24 g \( \mathrm{CaCN}_2\)

First, convert grams of \( \mathrm{CaCN}_2 \) to moles:\(n =\dfrac{5.24 \, \mathrm{g}}{80.10 \, \mathrm{g/mol}} = 0.065 \,\mathrm{mol} \ \mathrm{CaCN}_2 \)Next, use the mole ratio from the balanced equation \(2 \mathrm{mol} \ \mathrm{NH}_3 \ /1 \mathrm{mol} \ \mathrm{CaCN}_2\) to find moles of \(\mathrm{NH}_3\):\(\mathrm{moles} \ \mathrm{ \mathrm{NH}_3} = 0.065 \, \mathrm{mol} \ \mathrm{CaCN}_2 \times 2 \mathrm{mol} \ \mathrm{NH}_3 / \ \mathrm{mol} \ \mathrm{CaCN}_2 = 0.13 \, \mathrm{mol} \ \mathrm{NH}_3\)Finally, convert moles of \(\mathrm{NH}_3 \) to grams:\(\mathrm{grams} \ \mathrm{NH}_3 = 0.13 \, \mathrm{mol} \ \mathrm{NH}_3 \times 17.04 \, \mathrm{g/mol} = 2.22 \, \mathrm{g} \ \mathrm{NH}_3\)
05

Step 3c: Calculate grams of \( \mathrm{CaCO}_3\) formed if 155 g \( \mathrm{H}_2 \mathrm{O}\ \) reacts

Convert grams of \(\mathrm{H}_2O\) to moles:\(\mathrm{moles}\ \mathrm{H}_2O = \frac{155 \, \mathrm{g}}{18.02 \mathrm{{g/mol}}} = 8.60 \, \mathrm{mol} \ \mathrm{H}_2O\)Next, using the mole ratio from the balanced equation \(1 \ \mathrm{CaCN}_2\ / 3 \, \mathrm{H}_2O\), convert moles of \(\mathrm{H}_2O \) to moles of \(\mathrm{CaCN}_2\): \(\mathrm{moles} \ \mathrm{CaCN}_2 = 8.60 \, \mathrm{mol} \ \ \mathrm{H}_2O \times \frac{1 \, \mathrm{mol} \ \ \mathrm{CaCN}_2}{3 \mathrm{mol} \ \mathrm{H}_2O } = 2.87 \, \mathrm{mol} \ \ \mathrm{CaCN}_2 \)Finally, use the ratio (1 \ \ \mathrm{CaCO}_3/1 \ \ \mathrm{CaCN}_2) to find the amount of \ \ \mathrm{CaCO}_3 : \(\mathrm{grams} \ \ \mathrm{CaCO}_3 = 2.87 \, \mathrm{mol} \, \times \,100.09 \, \mathrm{g/mol} \ \mathrm{CaCO}_3 = 287 \, \mathrm{g} \ \ \mathrm{CaCO}_3\)

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Key Concepts

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

Stoichiometry
Stoichiometry is the calculation of reactants and products in chemical reactions. It鈥檚 all about determining how much of a substance is needed or produced. This involves using ratios from balanced chemical equations.
For example, in the given reaction, knowing that 1 mole of \(\text{CaCN}_{2}\) reacts with 3 moles of \(\text{H}_{2}O\), allows us to determine the quantities of each substance involved. It also helps us find out how much product will be formed.
To use stoichiometry effectively:
  • Start with a balanced chemical equation.
  • Identify the ratio of moles of reactants to products.
  • Use this ratio to relate the amounts of different substances.
Practicing stoichiometry helps in understanding reactions better and solving related problems becomes easier.
Mole Calculations
Moles are a fundamental unit in chemistry that represent a specific number of particles, usually atoms or molecules. One mole equals Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles.
In mole calculations, you often convert between the mass of a substance and the number of moles. The formula is:
\[ n = \frac{mass}{molecular \, weight} \]
For example, to calculate the moles of \(\text{CaCN}_{2}\) from 75.0 grams, you divide the mass by its molecular weight (80.10 g/mol):
\[ n = \frac{75.0 \, g}{80.10 \, g/mol} = 0.936 \, mol \]
This lets you translate between grams and moles, helping in the stoichiometric calculations.
Balanced Chemical Equations
A balanced chemical equation has the same number of each type of atom on both sides of the equation. This balance maintains the law of conservation of mass.
The balanced equation for the reaction in the exercise is: \[ \text{CaCN}_{2}(s) + 3 \text{H}_{2}O(l) \rightarrow \text{CaCO}_{3}(s) + 2 \text{NH}_{3}(g) \]
To balance a chemical equation:
  • Write down the number of atoms for each element on both sides.
  • Adjust coefficients to get the same number of atoms for each element on both sides.
  • Ensure the smallest possible whole number coefficients.
A balanced equation is crucial for accurate stoichiometric calculations, as it provides the mole ratios needed.
Molecular Weight Calculation
Molecular weight (or molar mass) is the mass of one mole of a substance, measured in grams per mole (g/mol). It is the sum of the atomic weights of all the atoms in the molecule.
To calculate the molecular weight, use the atomic masses from the periodic table:
  • \text{CaCN}_{2}: \[40.08 \, (Ca) + 2(12.01 \, (C) + 14.01 \, (N)) = 80.10 \, g/mol \]
  • \text{H}_{2}O: \[2(1.01) + 16.00 = 18.02 \, g/mol \]
  • \text{NH}_{3}: \[14.01 + 3(1.01) = 17.04 \, g/mol \]
  • \text{CaCO}_{3}: \[40.08 + 12.01 + 3(16.00) = 100.09 \, g/mol \]
Knowing the molecular weight allows you to convert between the mass of a substance and moles, making it essential for stoichiometric calculations.

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Most popular questions from this chapter

Calculate the enthalpy change for the reaction \(\mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)\) from the following: $$ \begin{array}{ll} \mathrm{S}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{3}(g) & \Delta H=-396 \mathrm{~kJ} \\ \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{3}(g) & \Delta H=-90 \mathrm{~kJ} \end{array} $$

For each of the following reactions, calculate the grams of indicated product when \(15.0 \mathrm{~g}\) of the first reactant and \(10.0 \mathrm{~g}\) of the second reactant are used: a. \(4 \mathrm{Li}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Li}_{2} \mathrm{O}(s) \quad\left(\mathrm{Li}_{2} \mathrm{O}\right)\) b. \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(l) \quad(\mathrm{Fe})\) c. \(\mathrm{Al}_{2} \mathrm{~S}_{3}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) $$ 2 \mathrm{Al}(\mathrm{OH})_{3}(a q)+3 \mathrm{H}_{2} \mathrm{~S}(g) \quad\left(\mathrm{H}_{2} \mathrm{~S}\right) $$

In an endothermic reaction, is the energy of the products higher or lower than that of the reactants?

Classify each of the following as exothermic or endothermic and give the \(\Delta H\) for each: a. \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)+802 \mathrm{~kJ}\) b. \(\mathrm{Ca}(\mathrm{OH})_{2}(s)+65.3 \mathrm{~kJ} \longrightarrow \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) c. \(2 \mathrm{Al}(s)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Fe}(l)+850 \mathrm{~kJ}\)

Each of the following is a reaction that occurs in the cells of the body. Identify which is exothermic and endothermic. (9.6) a. Phosphocreatine \(+\mathrm{H}_{2} \mathrm{O} \longrightarrow\) creatine \(+\mathrm{P}_{\mathrm{i}}+42.7 \mathrm{~kJ}\) b. Fructose-6-phosphate \(+\mathrm{P}_{\mathrm{i}}+16 \mathrm{~kJ} \longrightarrow\) fructose- 1,6 -bisphosphate
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