64.0k views
1 vote
How many grams of magnesium metal are needed to produce 35.2 L of H2 gas at STP? Mg + 2HCl → H2 + MgCl2

1 Answer

3 votes

Molar volume at STP can be used to convert from moles to gas volume and from gas volume to moles.

At STP: 1 mol = 22.4 L

So we are given the volume of H2 = 35.2 L

We can use that volume to find number of moles, then from number of moles we can find the mass of H2.

1 mol = 22.4 L

x mol = 35.2 L

22.4 L. x mol= 35.2 L x 1 mol

x = 35.2/22.4

x = 1.571 mol

H2 have 1.571 number of moles.

Since we want the mass of Mg, we will use the stoichiometry, for every 1 mole of Mg, 1 mole of H2 is produced. So the molar ration is 1:1. Therefore the number of moles of Mg = 1.571 mol

We can use this equation to find the mass:

n = m/M where n is the number of moles, m is the mass and M is the molar mass of Mg.

m = n x M

m = 1.571 x 24,305

m = 38.183 g

User Ricmarchao
by
4.1k points