In the reaction Mg (s) + 2HCl (aq) H2 (g) + MgCl2 (aq), how many moles of hydrogen gas will be produced from 75.0 milliliters of a 1.0 M HCl in an excess of Mg?

Question

In the reaction Mg (s) + 2HCl (aq)   H2 (g) + MgCl2 (aq), how many moles of hydrogen gas will be produced from 75.0 milliliters of a 1.0 M HCl in an excess of Mg?

anonymous · Answer

At STP, one mole of an ideal gas will occupy 22.414 L
Atmospheric hydrogen (H2 gas) has a density of 0.08988 g/L
[Because elemental hydrogen (H) has a molar mass of approx 1 g/mol so if you take 1 mole of atmospheric hydrogen (M is approx 2 g/mol) and divide the mass (2g) by 22.414L you will get approx 0.08988 g/L]

Then just multiply 0.08988 g/L by the volume in liters (75mL=0.075L) you will get the mass, divide this by the molar mass of atmospheric hydrogen and you will get the number of moles of evolved atmospheric hydrogen.

$$0.08988g/L \times \frac{ 75 }{ 1000 }L = 0.006741g$$
\[\frac{ 0.006741g }{ 2g/mol } = 0.0033705 mol = 3.3705 \times 10^{-3} moles