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Chapter 4: Problem 64

The reaction of calcium hydride and water produces calcium hydroxide and hydrogen as products. How many moles of \(\mathrm{H}_{2}(\mathrm{g})\) will be formed in the reaction between \(0.82 \mathrm{mol} \mathrm{CaH}_{2}(\mathrm{s})\) and \(1.54 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ?\)

Short Answer

Expert verified
1.54 moles of hydrogen gas will be produced.

Step by step solution

01

Analyze the stoichiometry of the reaction

It is clear from the balanced equation that for every mole of \(\mathrm{CaH}_{2}\), two moles of \(\mathrm{H}_{2}\) are generated.
02

Find the number of moles of hydrogen produced

Using stoichiometry, since we start with 0.82 moles of \(\mathrm{CaH}_{2}\), we will produce 0.82 moles 脳 2 = 1.64 moles of \(\mathrm{H}_{2}\) gas.
03

Compare with available moles of water

Ensure there is enough \(\mathrm{H}_{2}\mathrm{O}\) to react with the \(\mathrm{CaH}_{2}\). This is not limiting as we have 1.54 moles of \(\mathrm{H}_{2}\mathrm{O}\), and for every mole of \(\mathrm{CaH}_{2}\), only 2 moles of \(\mathrm{H}_{2}\mathrm{O}\) are required. 1.54 is less than 2脳0.82, so water limits the reaction.
04

Calculate the number of moles of hydrogen produced when water limits the reaction

From every 2 moles of water, 2 moles of \(\mathrm{H}_{2}\) gas are produced. Hence, the amount of \(\mathrm{H}_{2}\) gas that is produced when water limits the reaction is 1.54 moles.

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Key Concepts

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

Reaction Stoichiometry
When discussing reaction stoichiometry, we are focusing on the relationships between the quantities of reactants and products involved in a chemical reaction. In the given reaction between calcium hydride (\(\text{CaH}_2\)) and water (\(\text{H}_2\text{O}\)) to form calcium hydroxide (\(\text{Ca(OH)}_2\)) and hydrogen gas (\(\text{H}_2\)), stoichiometry helps us understand these relationships.
The balanced chemical equation for this reaction is:\[\text{CaH}_2(s) + 2 \text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(aq) + 2 \text{H}_2(g)\]
  • This indicates that 1 mole of calcium hydride reacts with 2 moles of water.
  • The reaction produces 1 mole of calcium hydroxide and 2 moles of hydrogen gas.
This proportion is crucial for determining how much product can be formed from given amounts of reactants. Stoichiometry is used in chemistry to calculate and predict the outcomes of chemical reactions.
Limiting Reactant
The concept of a limiting reactant is essential in determining how much product can be formed in a reaction. In any chemical reaction, the limiting reactant is the substance that is totally consumed first, thereby limiting the amount of products formed.
In the exercise, we started with 0.82 moles of calcium hydride and 1.54 moles of water. The balanced chemical equation tells us that every mole of calcium hydride requires 2 moles of water to react completely.
  • Thus, for 0.82 moles of calcium hydride, 0.82 \(\times\) 2 = 1.64 moles of water are needed.
  • We only have 1.54 moles of water available.
Since we do not have the required 1.64 moles of water, water is our limiting reactant.
This concept explains why, even though we have enough calcium hydride to form more hydrogen, the reaction will stop once the 1.54 moles of water are used up. Knowing the limiting reactant helps us accurately determine the amount of product formed.
Chemical Equation
A chemical equation is a symbolic representation of a chemical reaction. It shows the reactants transforming into products and provides useful information about the quantities and identities of the substances involved.
For the reaction between calcium hydride and water:\[\text{CaH}_2(s) + 2 \text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(aq) + 2 \text{H}_2(g)\]
  • The symbols and formulas tell us which substances are present and in what state they are (solid, liquid, gas, etc.).
  • The coefficients (numbers in front) indicate how many moles of each substance are involved.
Balancing a chemical equation is a key skill in chemistry. It ensures that the same number of each type of atom appears on both sides of the reaction, following the law of conservation of mass. This balance allows chemists to predict the outcomes of reactions and to calculate amounts using stoichiometry with confidence.

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Most popular questions from this chapter

How many grams of sodium must react with \(155 \mathrm{mL}\) \(\mathrm{H}_{2} \mathrm{O}\) to produce a solution that is \(0.175 \mathrm{M} \mathrm{NaOH} ?\) (Assume a final solution volume of \(155 \mathrm{mL}\) ) $$ 2 \mathrm{Na}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{NaOH}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g}) $$

A 0.696 mol sample of \(\mathrm{Cu}\) is added to \(136 \mathrm{mL}\) of \(6.0 \mathrm{M}\) HNO \(_{3}\). Assuming the following reaction is the only one that occurs, will the Cu react completely? $$\begin{aligned} 3 \mathrm{Cu}(\mathrm{s})+8 \mathrm{HNO}_{3}(\mathrm{aq}) & \longrightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) +4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+2 \mathrm{NO}(\mathrm{g}) \end{aligned}$$

Which has the higher concentration of sucrose: a \(46 \%\) sucrose solution by mass \((d=1.21 \mathrm{g} / \mathrm{mL}),\) or \(1.50 \mathrm{M}\) \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) ? Explain your reasoning.

Ammonia can be generated by heating together the solids \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(\mathrm{Ca}(\mathrm{OH})_{2} . \mathrm{CaCl}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are also formed. (a) If a mixture containing \(33.0 \mathrm{g}\) each of \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(\mathrm{Ca}(\mathrm{OH})_{2}\) is heated, how many grams of \(\mathrm{NH}_{3}\) will form? (b) Which reactant remains in excess, and in what mass?

Chalkboard chalk is made from calcium carbonate and calcium sulfate, with minor impurities such as \(\mathrm{SiO}_{2} .\) Only the \(\mathrm{CaCO}_{3}\) reacts with dilute \(\mathrm{HCl}(\mathrm{aq})\) What is the mass percent \(\mathrm{CaCO}_{3}\) in a piece of chalk if a 3.28 -g sample yields \(0.981 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g}) ?\) $$\begin{aligned} \mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) & \longrightarrow \mathrm{CaCl}_{2}(\mathrm{aq}) +\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) \end{aligned}$$
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