Question

Exercise 11.105---- Part A

The reaction between zinc and hydrochloric acid is carried out as a source of hydrogen gas in the laboratory:
Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g)

If 321 mL of hydrogen gas is collected over water at 25∘C at a total pressure of 746 mmHg , how many grams of Zn reacted?

Exercise 11.106--- Part B

Consider the reaction:
2NiO(s)→2Ni(s)+O2(g)

If O2 is collected over water at 40 ∘C and a total pressure of 747 mmHg , what volume of gas will be collected for the complete reaction of 25.51 g of NiO?

Exercise 11.107---- Part C

How many grams of hydrogen are collected in a reaction where 1.84 L of hydrogen gas is collected over water at a temperature of 40 ∘C and a total pressure of 747 torr ?

Answer 1

Part A

water vapour pressure at 25⁰C = 23.8 mmHg

hydrogen pressure = 746 - 23.8 = 722.2 mmHg

by using ideal gas equation calculate no. of mole of hydrogen collected

PV = nRT             where, P = atm pressure= 722.2 mmHg = 0.95 atm,

V = volume in Liter = 321 ml = 0.321 L

n = number of mole = ?

R = 0.08205L atm mol^-1 K^-1 =Proportionality constant = gas constant,

T = Temperature in K = 25⁰C = 273.15+ 25 = 298.15 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (0.95X 0.321)/(0.08205X 298.15) = 0.01247 mole

0.01247 mole of H₂ gas produced

According to reaction 1 mole of Zn produce 1 mole of H₂ then to produce 0.01247 mole of H₂ required Zn = 0.01247 mole

molar mass of Zn = 65.38 gm/mole

then 0.01247 mole of Zn = 0.01247 X 65.38 = 0.815 gm

0.815 gm of Zn reacted