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

The answer is !

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