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Chapter 9: Problem 86

When small quantities of elemental hydrogen gas are needed for laboratory work, the hydrogen is often generated by chemical reaction of a metal with acid. For example, zinc reacts with hydrochloric acid, releasing gaseous elemental hydrogen: $$ \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ What mass of hydrogen gas is produced when \(2.50 \mathrm{~g}\) of zine is reacted with excess aqueous hydrochloric acid?

Short Answer

Expert verified
When \(2.50 \ \text{g}\) of zinc reacts with excess aqueous hydrochloric acid, \(0.0771 \ \text{g}\) of hydrogen gas is produced.

Step by step solution

01

Determine the molar mass of the substances involved#important# In order to perform stoichiometric calculations, we need to know the molar mass of zinc and hydrogen: - Molar mass of zinc (Zn): 65.38 g/mol - Molar mass of hydrogen (H鈧�): 2.02 g/mol

Step 2: Calculate moles of zinc reacted#important# To find the moles of zinc reacted, we can use the given mass (2.50 g) and the molar mass of zinc (65.38 g/mol): $$ \text{moles of Zn} = \frac{\text{mass of Zn}}{\text{molar mass of Zn}} = \frac{2.50 \ \text{g}}{65.38 \ \text{g/mol}} = 0.0382 \ \text{mol} $$
02

Calculate moles of hydrogen produced#important# The stoichiometric ratio between Zn and H鈧� from the balanced equation is 1:1. Therefore, the moles of hydrogen produced would be the same as the moles of zinc reacted: $$ \text{moles of H鈧倉 = \text{moles of Zn} = 0.0382 \ \text{mol} $$

Step 4: Calculate mass of hydrogen produced#important# Finally, we can calculate the mass of hydrogen gas produced by using the calculated moles of hydrogen and its molar mass (2.02 g/mol): $$ \text{mass of H鈧倉 = \text{moles of H鈧倉 \times \text{molar mass of H鈧倉 = 0.0382 \ \text{mol} \times 2.02 \ \text{g/mol} = 0.0771 \ \text{g} $$ So, when 2.50 g of zinc reacts with excess aqueous hydrochloric acid, 0.0771 g of hydrogen gas is produced.

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Key Concepts

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

Chemical Reactions
Chemical reactions describe the process by which substances interact and transform into new products. In this exercise, the reaction involves zinc (Zn) and hydrochloric acid (HCl). This is a single displacement reaction, a type where an element displaces another in a compound, leading to the formation of a new product.
  • Zinc, a metal, and hydrochloric acid, an aqueous solution, react to form zinc chloride (ZnCl2) and hydrogen gas (H2).
  • The balanced chemical equation for this reaction is: \( \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) \).
Chemical reactions like this one can be explored further using stoichiometry, which helps in predicting the amounts of products formed and reactants consumed.
Molar Mass Calculation
Molar mass is fundamental in stoichiometry, converting between mass and moles. It's the mass of one mole of a substance, expressed in grams per mole (g/mol). This concept is used to relate the mass of a chemical substance to the quantity in moles.
  • To calculate the molar mass of zinc, we use its atomic mass of 65.38 g/mol.
  • For hydrogen gas (H2), with two hydrogen atoms, its molar mass is 2.02 g/mol.
The calculation of moles from mass involves the equation: \[\text{moles} = \frac{\text{mass of substance}}{\text{molar mass}}\]This conversion is critical in predicting product amounts by using the balanced reaction equation as a guide.
Hydrogen Gas Production
In this type of reaction, understanding how to calculate and predict hydrogen gas yield is crucial. Producing hydrogen gas involves using metals like zinc reacting with acids, which is a common laboratory practice.
  • Starting with 2.50 g of zinc, we determine moles using its molar mass: \( 0.0382 \text{ mol of Zn} \).
  • The stoichiometry of the reaction shows a 1:1 mole ratio between zinc and hydrogen gas.
  • This means the moles of hydrogen gas produced equals the moles of zinc reacted: \( 0.0382 \text{ mol of H}_2 \).
  • Finally, converting moles of hydrogen to mass involves multiplying by the molar mass of hydrogen, resulting in 0.0771 g of hydrogen gas.
Such calculations underline the importance of understanding stoichiometry and reactions to estimate gas production efficiently.

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Most popular questions from this chapter

Small quantities of ammonia gas can be generated in the laboratory by heating an ammonium salt with a strong base. For example, ammonium chloride reacts with sodium hydroxide according to the following balanced equation: $$ \mathrm{NH}_{4} \mathrm{Cl}(s)+\mathrm{NaOH}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{NaCl}(s)+\mathrm{H}_{2} \mathrm{O}(g) $$ What mass of ammonia gas is produced if \(1.39 \mathrm{~g}\) of ammonium chloride reacts completely?

Using the average atomic masses given inside the front cover of the text, calculate how many moles of each substance the following masses represent. a. 4.21 g of copper(II) sulfate b. \(7.94 \mathrm{~g}\) of barium nitrate c. \(1.24 \mathrm{mg}\) of water d. \(9.79 \mathrm{~g}\) of tungsten c. 1.45 lb of sulfur f. 4.65 g of ethyl alcohol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) g. 12.01 g of carbon

For each of the following unbalanced chemical 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. \(\mathrm{CS}_{2}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{SO}_{2}(g)\) b. \(\mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{CN}_{2} \mathrm{H}_{4} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{MnO}_{2}(s) \rightarrow \mathrm{MnO}(s)+\mathbf{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{I}_{2}(l)+\mathrm{Cl}_{2}(g) \rightarrow \mathbf{I C l}(g)\)

Explain how one determines which reactant in a process is the limiting reactant. Does this depend only on the masses of the reactant present? Give an example of how to determine the limiting reactant by using a Before-Change- After (BCA) table with a balanced chemical cquation and reactant starting amounts.

For each of the following unbalanced reactions, suppose exactly 5.00 moles of each reactant are taken. Determine which reactant is limiting, and also determine what mass of the excess reagent will remain after the limiting reactant is consumed. For cach reaction, solve the problem three ways: i. Set up and use Before-Change-After (BCA) tables. ii. Compare the moles of reactants to see which runs out first. iii. Consider the amounts of products that can be formed by completcly consuming cach reactant. a. \(\mathrm{CaC}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g)\) b. \(\operatorname{AgNO}_{3}(a q)+\mathbf{A l}(s) \rightarrow \mathbf{A}_{\mathbf{g}}(s)+\mathbf{A l}\left(\mathrm{NO}_{3}\right)_{3}(a q)\)
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