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

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.