Question

When 15.0 g of zinc metal reacts with excess hcl, how many liters of h2 gas are produced at stp?

Answer 1

To find the volume of H2 gas produced when 15.0 g of zinc reacts with excess HCl at STP, we first calculate the moles of Zn from its mass and molar mass. Using the stoichiometry of the reaction, we know that the moles of zinc equal the moles of H2 produced. Multiplying the moles of H2 by the molar volume at STP gives us 5.1416 L of H2 gas.

Step-by-step explanation:

When 15.0 g of zinc metal reacts with excess HCl, the amount of H2 gas produced at STP (Standard Temperature and Pressure) can be calculated using the stoichiometry of the balanced chemical equation and the ideal gas law. First, we need to write the balanced chemical equation:

Zn(s) + 2 HCl(aq) → ZnCl2 (aq) + H2(g)

According to the equation, one mole of Zn reacts with two moles of HCl to produce one mole of H2 gas. We convert the mass of Zn to moles by using its molar mass (65.38 g/mol):

15.0 g Zn × (1 mol Zn / 65.38 g Zn) = 0.2295 mol Zn

Then, since one mole of Zn produces one mole of H2, we have 0.2295 moles of H2. At STP, one mole of any gas occupies 22.4 L, so:

0.2295 mol H2 × (22.4 L / mol) = 5.1416 L of H2

Answer 2

M(Zn)=65.4 g/mol
15.0g Zn*1mol Zn/65.4 g Zn = 15/65.4 mol Zn

Zn +2HCl--->ZnCl2 + H2
by equation 1 mol 1 mol
from the problem 15/65.4 mol 15/65.4 mol

Molar volume of any gas under STP = 22.4 L/mol

15/65.4 mol H2 * 22.4 L/1 mol= 5.14 L H2