Question

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

A) What volume of hydrogen at 0 ∘C and 1.00 atm (STP) is released when 6.20 g of Mg reacts?

B) How many grams of magnesium are needed to prepare 4.85 L of H2 at 735 mmHg and 21 ∘C?

Answer 1

A)

Molar mass of Mg = 24.31 g/mol

mass of Mg = 6.2 g

mol of Mg = (mass)/(molar mass)

= 6.2/24.31

= 0.255 mol

According to balanced equation

mol of H2 formed = moles of Mg

= 0.255 mol

Given:

P = 1.0 atm

n = 0.255 mol

T = 0.0 oC

= (0.0+273) K

= 273 K

use:

P * V = n*R*T

1 atm * V = 0.255 mol* 0.08206 atm.L/mol.K * 273 K

V = 5.7126 L

Answer: 5.71 L

B)

Given:

P = 735.0 mm Hg

= (735.0/760) atm

= 0.9671 atm

V = 4.85 L

T = 21.0 oC

= (21.0+273) K

= 294 K

find number of moles using:

P * V = n*R*T

0.9671 atm * 4.85 L = n * 0.08206 atm.L/mol.K * 294 K

n = 0.1944 mol

From reaction,

Mol of Mg reacted = mol of H2 produced

= 0.1944 mol

Molar mass of Mg = 24.31 g/mol

use:

mass of Mg,

m = number of mol * molar mass

= 0.1944 mol * 24.31 g/mol

= 4.726 g

Answer: 4.73 g