Question

What volume of hydrogen gas at STP will be produced from 36.7 g of magnesium reacting with an excess amount of hydrochloric acid?

Answer 1

V = 34.3 L

Step-by-step explanation:

Given data:

Volume of hydrogen produced = ?

Mass of magnesium = 36.7 g

Temperature and pressure = standard

Solution:

Mg + 2HCl → MgCl₂ + H₂

Number of moles of magnesium:

Number of moles = mass/molar mass

Number of moles = 36.7 g/ 24 g/mol

Number of moles = 1.53 mol

Now we will compare the moles of magnesium with hydrogen.

Mg : H₂

1 : 1

1.53 : 1.53

Number of moles of hydrogen produced = 1.53 mol

Volume of hydrogen:

PV = nRT

V = nRT / P

V = 1.53 mol × 0.0821 atm.L/mol.K × 273 K/ 1 atm

V = 34.3 L