In this problem, zinc reacts with hydrochloric acid to make zinc chloride and hydrogen gas according to the following reaction: Zn(s) + HCl(aq) → ZnCl2(aq) + H2(g) a. Balance the reaction b. A 3.50 g sample of zinc metal is allowed to react with 2.50 g of hydrochloric acid. Show your work below and complete the following table: Reactants and Products Before Reaction After Reaction Zn (grams) 3.50 g 2.24 g HCl (grams) 2.50 g 0 g ZnCl2 (grams) 0 g 2.24 g H2 (L) 0 L 0.56 L
The Correct Answer and Explanation is:
Corrected Answer
a. Balanced Reaction
Zn(s) + 2 HCl(aq) → ZnCl₂(aq) + H₂(g)
b. Final State of the Reaction
The following table shows the correct quantities of reactants and products before and after the reaction is complete.
| Reactants and Products | Before Reaction | After Reaction |
| Zn (grams) | 3.50 g | 1.26 g |
| HCl (grams) | 2.50 g | 0 g |
| ZnCl₂ (grams) | 0 g | 4.67 g |
| H₂ (L) | 0 L | 0.768 L |
Explanation
This problem is an example of a limiting reactant calculation in stoichiometry. The amounts of all substances after the reaction depend on which reactant is completely consumed first. The process involves converting the initial masses of reactants into moles, using the balanced chemical equation to determine the limiting reactant, and then calculating the quantities of products formed and excess reactant remaining.
Step 1: Balancing the Chemical Equation
The initial unbalanced equation is Zn + HCl → ZnCl₂ + H₂. To balance the equation, the number of atoms of each element must be the same on both the reactant and product sides. There is one zinc atom on both sides. However, there is one hydrogen and one chlorine atom on the reactant side, but two of each on the product side. Placing a coefficient of 2 in front of HCl balances the hydrogen and chlorine atoms.
The balanced equation is: Zn(s) + 2 HCl(aq) → ZnCl₂(aq) + H₂(g)
This equation shows that one mole of solid zinc reacts with two moles of hydrochloric acid to produce one mole of zinc chloride and one mole of hydrogen gas.
Step 2: Calculating Initial Moles of Reactants
To determine the limiting reactant, the initial masses must be converted to moles using their respective molar masses.
- Molar mass of Zn = 65.38 g/mol
- Molar mass of HCl = 36.46 g/mol (1.01 for H + 35.45 for Cl)
Moles of Zn = 3.50 g Zn / 65.38 g/mol = 0.0535 mol Zn
Moles of HCl = 2.50 g HCl / 36.46 g/mol = 0.0686 mol HCl
Step 3: Identifying the Limiting Reactant
The limiting reactant is the one that produces the least amount of product. The amount of product formed from each reactant can be calculated using the mole ratios from the balanced equation. Let’s calculate the moles of H₂ that could be produced from each reactant.
- From Zn: 0.0535 mol Zn × (1 mol H₂ / 1 mol Zn) = 0.0535 mol H₂
- From HCl: 0.0686 mol HCl × (1 mol H₂ / 2 mol HCl) = 0.0343 mol H₂
Since hydrochloric acid produces a smaller amount of hydrogen gas (0.0343 mol), HCl is the limiting reactant. It will be completely consumed, and the reaction will stop once all the HCl has reacted.
Step 4: Calculating Final Quantities
All subsequent calculations are based on the amount of the limiting reactant (0.0686 mol HCl).
- Mass of ZnCl₂ Produced:
First, find the moles of ZnCl₂ produced.
Moles of ZnCl₂ = 0.0686 mol HCl × (1 mol ZnCl₂ / 2 mol HCl) = 0.0343 mol ZnCl₂
Next, convert moles to grams using the molar mass of ZnCl₂ (136.28 g/mol ).
Mass of ZnCl₂ = 0.0343 mol × 136.28 g/mol = 4.67 g ZnCl₂ - Volume of H₂ Produced:
The moles of H₂ produced are 0.0343 mol. Assuming Standard Temperature and Pressure (STP), one mole of any ideal gas occupies 22.4 liters.
Volume of H₂ = 0.0343 mol × 22.4 L/mol = 0.768 L H₂ - Mass of Zn Remaining (Excess Reactant):
First, calculate the mass of Zn that reacted with the limiting HCl.
Moles of Zn reacted = 0.0686 mol HCl × (1 mol Zn / 2 mol HCl) = 0.0343 mol Zn
Mass of Zn reacted = 0.0343 mol × 65.38 g/mol = 2.24 g Zn
Finally, subtract the reacted mass from the initial mass.
Mass of Zn remaining = 3.50 g (initial) – 2.24 g (reacted) = 1.26 g Zn - Mass of HCl Remaining:
Since HCl is the limiting reactant, it is completely consumed.
Mass of HCl remaining = 0 g
