Question

Mg metal reacts with HCl to produce hydrogen gas.
Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)

What volume of hydrogen at 0 ∘C and 1.00 atm (STP) is released when 8.80 g of Mg reacts?

Express your answer with the appropriate units.

How many grams of magnesium are needed to prepare 5.75 L of H2 at 745 mmHg and 16 ∘C?

Express your answer with the appropriate units.

Answer 1

Mg_(s)   + 2HCl_(aq) --------------> MgCl2_(aq) + H2_(g)

a) 24.3 g Mg gives 1 mole H2

8.80 g Mg gives 8.80 x 1 / 24.3 = 0.362 moles H2

1mole = 22.4 L at STP

0.362 moles H2 = 0.362 x 22.4 / 1 = 8.11 L

volume of H2 released = 8.11 L

b) calculate the moles of H2

PV = nRT

n = PV / RT

P = 745 mmHg = 0.98 atm

T = 16⁰C = 289 K

n = (0.98 x 5.75 / 0.0821 x 289)

n = 5.635 / 23.7269

n = 0.237

1 mole H2 require 24.3 g Mg

0.237 moles H2 require 0.237 x 24.3 / 1 = 5.759 g

mass of Mg reacted = 5.759 g