# In the reaction of aluminum with sulfuric acid, 3.42 g of aluminum sulfate was formed.

**In the reaction of aluminum with sulfuric acid, 3.42 g of aluminum sulfate was formed. Determine the mass and amount (mol) of aluminum that has reacted.**

1.Calculate the amount of aluminum sulfate substance by the formula:

n = m: M, where M is molar mass.

M (Al2 (SO4) 3) = 27 × 2 + 3 × (32 + 64) = 342 g / mol.

n = 3.42: 342 g / mol = 0.01 mol.

2. Let’s compose the reaction equation:

2Al + 3H2SO4 = Al2 (SO4) 3+ 3H2.

According to the reaction equation, 2 moles of aluminum react with 1 mole of aluminum sulfate, that is, the substances are in quantitative ratios of 2: 1, which means the amount of aluminum substance will be 2 times more than the amount of aluminum sulfate substance:

n (Al) = 0.01 × 2 = 0.02 mol.

2 mol Al – 1 mol Al2 (SO4) 3,

n mol Al – 0.01 mol Al2 (SO4) 3.

n mol = (0.01 mol × 2 mol): 1 = 0.02 mol.

3.Calculate the mass of aluminum by the formula:

m = n M,

M (Al) = 27 g / mol.

m = 27 g / mol × 0.02 mol = 0.54 g.

Answer: m = 0.54 g.