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.