Question

Mg metal reacts with HCl to produce hydrogen gas. Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g) Part A What volume of hydrogen at 0 ∘C and 1.00 atm (STP) is released when 9.45 g of Mg reacts? Express your answer with the appropriate units.

Answer 1

Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)

(since the molar mass of Mg is not given, assuming that it's mola mass is 24gmol¯1)

1st find the moles of Mg using the equation n=m/M where

n - moles

m - mass

M - molar mass

Therefore :

$n \: = 9.45 \:g\div 24g {mol}^{ - 1}$

n = 0.39375mol

n = 0.39mol

Then using the equation n=V/Vm where;

n = mol

V = volume

Vm = molar volume

Find the volume.

n = V/Vm

(n =0.39mol , Vm = 22.4dm³mol¯¹)

V = 0.39mol×22.4dm³mol¯¹

V = 8.736dm³