If 35.2 g of hydrochloric acid react with excess of magnesium, how many grams of hydrogen are produced?

Mg + 2HCl - -> MgCl2 + H2

0.975 g of H₂

Explanation:

The equation for the reaction between magnesium and HCl is given by;

Mg + 2HCl - -> MgCl2 + H2

We are given 35.2 g of of HCl reacted with excess magnesium.

We are required to calculate the amount of Hydrogen produced;

Step 1: Calculation of number of moles of HCl

To calculate the number of moles we divide the mass by the molar mass

Molar mass of HCl = 36.46 g/mol

Mass of HCl = 35.2 g

Number of moles = 35.2 g : 36.46 g/mol

= 0.965 moles

Step 2: Moles of Hydrogen produced

From the equation, the mole ratio of HCl to H₂ is 2 : 1

Therefore; Moles of Hydrogen = Moles of HCl : 2

= 0.965 moles : 2

= 0.4825 moles

Step 3: Mass of hydrogen gas produced

Mass is calculated by multiplying the number of moles by the molar mass.

Molar mass of hydrogen = 2.02 g/mol

Moles of hydrogen = 0.4825 moles

Mass of Hydrogen = 0.4825 moles * 2.02 g/mol

= 0.97465

= 0.975 g

Therefore; 0.975 g of hydrogen will be produced.