Question

If 161 mL of wet H2 is collected over water at 24 ∘C and a barometric pressure of 739 torr, how many grams of Zn have been consumed? (The vapor pressure of water is 22.38 torr.) m m = .420 g

Answer 1

Answer: 0.407 grams of Zn has been consumed.

Explanation: Zn reacts with hydrochloric acid to form hydrogen gas. The balanced equation is given below:

$Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)$

Barometric pressure is given as 739 torr which is the sum of partial pressures of hydrogen gas and water vapors. Water vapor pressure is given as 22.38 torr.

So, partial pressure of dry hydrogen gas = 739 torr - 22.38 torr  = 716.62 torr

Convert the pressure from torr to atm.

760 torr = 1 atm

so, $716.62torr(\frac{1atm}{760torr})$

= 0.943 atm

Temperature is given as 24 degree C. In kelvin it will be 24+273 = 297 K

Volume of the gas is given as 161 mL that is 0.161 L.

From this information we can calculate the moles of hydrogen gas formed using ideal gas law equation.

PV = nRT

$n=\frac{PV}{RT}$

R is universal gas law constant and its value is 0.0821 atm.L/(mol.K).

Let's plug in the values in the equation to calculate moles(n) of the gas.

$n=\frac{0.943atm*0.161L}{0.0821atm*L*mol^-^1*K^-^1*297K}$

n = 0.00623 moles

From the balanced equation, there is 1:1 mol ratio between zinc and hydrogen gas. So, the moles of Zn used will also be 0.00623.

Multiply the moles by molar mass to convert into grams. Molar mass of Zn is 65.38 gram per mol.

$0.00623mol(\frac{65.38g}{1mol})$

= 0.407 gram

So, 0.407 grams of Zn have been consumed.