Question

Hydrogen gas is produced when zinc reacts with sulfuric acid:
Zn(s)+H2SO4(aq)→ZnSO4(aq)+H2(g)

If 155 mL of wet H2 is collected over water at 24 ∘C and a barometric pressure of 736 torr, how many grams of Zn have been consumed? (The vapor pressure of water is 22.38 torr.)

Answer 1

1st find the moles of H2 produced

we have:

P = 736 torr - 22.38 torr

= 713.62 torr

= (713.62/760) atm

= 0.939 atm

V = 155.0 mL

= (155.0/1000) L

= 0.155 L

T = 24.0 oC

= (24.0+273) K

= 297 K

find number of moles using:

P * V = n*R*T

0.939 atm * 0.155 L = n * 0.08205 atm.L/mol.K * 297 K

n = 5.972*10^-3 mol

from reaction,

moles of Zn reacted = moles of H2 produced

= 5.972*10^-3 mol

Molar mass of Zn = 65.38 g/mol

we have below equation to be used:

mass of Zn,

m = number of mol * molar mass

= 5.972*10^-3 mol * 65.38 g/mol

= 0.3904 g

Answer: 0.390 g