0.2 g of a magnesium-copper alloy containing 20% copper was treated with an excess of 10% sulfuric acid solution

0.2 g of a magnesium-copper alloy containing 20% copper was treated with an excess of 10% sulfuric acid solution at a temperature of 25 degrees and a pressure of 1 atm. Determine the amount of gas evolved.

There are two metals – Mg and Cu, which were affected by a solution of H2SO4 (n.y). But only magnesium will enter into the reaction (copper does not displace hydrogen from acids).

1. Let’s find the mass of magnesium. If the alloy contains 20% copper, then the proportion of magnesium will be 80%

0.2 g alloy – 100%

x g magnesium – 80%

Hence, x = 0.2 * 80/100 = 0.16 g Mg

2. Let’s make the equation:

Mg + H2SO4 = MgSO4 + H2

According to the equation: 1 mol of magnesium participates in the reaction

m (Mg) = n * M = 1 * 24 = 24g

According to the equation: 1 mole of hydrogen turned out, i.e. 22.4 l

We make the proportion:

0.16g Mg – xl H2

24g Mg – 22.4L H2

Hence, x = 0.16 * 22.4 / 24 = 0.15 liters of hydrogen.

There are two metals – Mg and Cu, which were affected by a solution of H2SO4 (n.y). But only magnesium will enter into the reaction (copper does not displace hydrogen from acids).

1. Let’s find the mass of magnesium. If the alloy contains 20% copper, then the proportion of magnesium will be 80%

0.2 g alloy – 100%

x g magnesium – 80%

Hence, x = 0.2 * 80/100 = 0.16 g Mg

2. Let’s make the equation:

Mg + H2SO4 = MgSO4 + H2

According to the equation: 1 mol of magnesium participates in the reaction

m (Mg) = n * M = 1 * 24 = 24g

According to the equation: 1 mole of hydrogen turned out, i.e. 22.4 l

We make the proportion:

0.16g Mg – xl H2

24g Mg – 22.4L H2

Hence, x = 0.16 * 22.4 / 24 = 0.15 liters of hydrogen.