Question

Given the following balanced reaction of hydrochloric acid and oxygen gas forming chlorine gas and water, how many grams of hydrochloric acid will be needed to form 335 grams of chlorine gas, assuming there is excess oxygen present? (To find the molar mass in the problem, use the periodic table and round the mass to the hundreds place for calculation.)

a)344g
b)788g
c)9.42g

Answer 1

The balanced reaction of Hcl with oxygen gas will be

4HCl + O2 ---> 2H2O + 2Cl2(g)

Thus from balanced equation it is clear that four moles of HCl will give two moles of Cl2 gas

Or

For 2 moles of Cl2 gas we need four moles of HCl

Molar mass of HCl = 36.5 g / mole , four moles = 4 X 36.5 = 146g

Molar mass of Cl2 = 71g /mole

Hence

2 moles = 2 X 71 = 142 g

Therefore

for 142g of Cl2 we need 146g of Hcl

for one gram of Cl2 we need = 146 / 142g of HCl

For 335g of Cl2 we need = 146 X 335 / 142 = 344.4 g of HCl

Answer 2

The ratio of moles of reactants to moles of products can be seen from the coefficients in a balanced equation. In our case 4 moles of hydrochloric acid reacts with one mole of oxygen to produce two moles of chlorine and water. So, the ratio of moles of hydrochloric acid to moles of chlorine is 2:1. To determine the number moles, divide the mass by the mass of one mole.
Cl2 = 2 * 35.45 = 70.9 grams
Number of moles = 335 ÷ 70.9
This is approximately 4.72 moles. The number of moles of hydrochloric acid is twice this number.

Mass of one mole = 1 + 35.46 = 36.45 grams
Total mass = 2 * (335 ÷ 70.9) * 36.45
This is approximately 344.45 grams.
Correct answer A.