How many liters of h2 gas, collected over water at an atmospheric pressure of 752 mmhg and a temperature of 21.0°c, can be made from 1.566 g of zn and excess hcl? the partial pressure of water vapor is 18.65 mmhg at 21.0°c?

Answer: 0.299 liter

Explanation:

1) Balanced chemical equation

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

2) Mole ratios

1 mol Zn (s) : 2 mol HCl (aq) : 1mol ZnCl₂ (aq) : 2 mol H₂ (g)

3) Convert 1.566 g Zn into number of moles (n)

n = 1.566g / 65.39 g/mol = 0.02395 mol

4) Set the proportion

1 mol Zn / 2 mol H₂ = 0.02395 mol Zn / x ⇒ x = 0.01197 mol H₂

5) Calcualte the partial pressure of H₂

Partial pressure H₂ = Total pressure - partial pressure of water vapor

Partial pressure H₂ = 752 mmHg - 18.65 mmHg = 733.35 mmHg

Convert to atm = 733.5 mmHg * 1 atm / 760mmHg = 0.9649 atm

6) Calculate volume using ideal gas equation

PV = nRT ⇒ V = nRT/P

⇒ V = 0.01197 mol * 0.0821 atm*liter / k*mol * (21 + 273k) / 0.9649atm

V = 0.2994 liter

The answer is V = 0.6 L

The explanation:

1 - First we will get P (H):

p (H) = p total - p (H2O)

= 752 mmHg - 18.65 mmHg

= 733 mmHg

= 733 mmHg X (1 atm / 760 atm) = 0.964 atm

2 - then when the reaction equation is:

Zn (s) + 2HCl (aq) - > H2 (g) + ZnCl2 (aq)

so, then 1 mole of Zn will produce 1 mole of H2.

so, we need to get first the mole of Zn:

mole Zn = mass / molar mass

= 1.566g / 65.38g/mol

= 0.02395mol

∴ 0.02395 mol of Zn will produce 0.02395 mol of H2

by using the ideal gas equation we can get the volume V:

when:

PV = nRT

when T is the temperature = 21 + 273 = 294 K

and R is the ideal gas constant = 0.0821 L atm / K mol

and n is the number of moles = 0.2395 mol

and P is the pressure = 0.964 atm

so, by substitution:

V = (0.02395 moles H2) (0.0821 L atm / K mole) (294 K) / (0.964 atm)

= 0.6 L