What mass of silicon oxide (IV) should be formed when burning 60 grams of silicon containing 5% impurities?

1.Let’s find the mass of impurities:

60 g – 100%,

m g – 5%,

m = (60 g × 5%): 100% = 3 g.

60 g – 3 g = 57 g (pure silicon)

Or the second way: 100% – 5% = 95% (pure silicon). Let us find the mass of silicon, denoting 60 g of substance for 100%, and the mass of silicon, which is 95% for m.

60 g – 100%,

m g -95%,

m (Si) = (60g × 95%): 100% = 57g.

2.Let’s find the amount of silicon substance by the formula:

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

M (Si) = 28 g / mol.

n = 57 g: 28 g / mol = 2.04 mol.

3. We make the equation of the reaction of combustion of silicon, we find the quantitative ratios of silicon Si and silicon oxide SiO2.

Si + O2 = SiO2.

1 mol of silicon, when burning in oxygen, forms 1 mol of silicon oxide, that is, they react 1: 1, the amount of silicon Si and silicon oxide SiO2 will be the same.

n (Si) = n (SiO2) = 2.04 mol.

4.Calculate the mass of silicon oxide SiO2 by the formula:

m = n M, where M (SiO2) = 28 + 16 × 2 = 60 g / mol.

m (SiO2) = 60 g / mol × 2.04 = 122.4 g.

Answer: m (SiO2) = 122.4 g.