If 8cm^3 of H2 reacts with an excess of Cl2, calculate how much of the HCL (g) is produced.

First, we need the balanced equation: H₂ + Cl₂ - - - > 2HCl

since not much information is given, I am assuming we are at STP and that 22.4 Liters = 1 mol

1) let's convert the volume to moles using the molar volume of a gas. also we need to convert the cm₃ to mL, then to Liters.

8 cm³ (1 ml / 1 cm³) (1 L / 1000 mL) (1 mol / 22.4 Liters) = 3.6x10⁻⁴ moles of H₂

2) let's use the mole ratio of the balanced equation to convert moles of H₂ to moles of HCl

3.6x10⁻⁴ mol H₂ (2 mol HCl / 1 mol H₂) = 7.1x10⁻⁴ mol HCl

3) lastly, we convert the moles of HCl to grams using the molar mass.

molar mass of HCl = 1.01 + 35.5 = 36.51 g/mol

7.1x10⁻⁴ mol HCl (36.51 g/mol) = 0.026 grams HCl