What volume of H₂ would be collected at 21.5^oC and a pressure of 695 torr if 5.25 g of zinc react according to the equation: Zn(s)+ 2HCl(aq)→ ZnCl₂(aq)+ H₂(g)

Question

Quizplus · Accepted Answer

First, we need to use stoichiometry to determine the amount of HCl that reacts with the given amount of zinc:

5.25 g Zn x (1 mol Zn/65.38 g Zn) x (2 mol HCl/1 mol Zn) x (36.46 g HCl/1 mol HCl) = 10.5 g HCl

So 10.5 g of HCl will react to produce hydrogen gas. To find the volume of hydrogen gas produced, we can use the ideal gas law:

PV = nRT

where P = 695 torr, V is the volume in liters that we want to find, n is the number of moles of hydrogen gas, R is the ideal gas constant, and T = 21.5 + 273 = 294.5 K. We can solve for n using the stoichiometry:

5.25 g Zn x (1 mol Zn/65.38 g Zn) x (1 mol H2/1 mol Zn) = 0.0801 mol H2

So we have 0.0801 mol of H2. Plugging in the values for P, n, R, and T, we get:

(695 torr) V = (0.0801 mol) (0.08206 L atm/K mol) (294.5 K)

V = (0.0801 mol) (0.08206 L atm/K mol) (294.5 K) / (695 torr)

V = 2.12 L

Therefore, the volume of hydrogen gas produced is 2.12 L, which is answer choice D.

What Volume of H2 Would Be Collected at 21

What Volume of H₂ Would Be Collected at 21