Question

Calculate how many grams of NaHCO3 and mL of 1 M HCl is required to make 80 mL of CO2 for the following reaction: NaHCO3 (s) + HCl (aq) -> NaCl (aq)+ H2O (l)+ CO2 (g)

hint:remember that 1 M HCl contains 1 mole per liter.

Answer 1

Given Reaction

NaHCO₃+ HCl--------------> NaCl + H₂O + CO₂

44g (1mol) of CO₂ corresponds to 22400 mL (22.4L at STP)

80mL of CO₂ corresponds to 3.57x10^-3mol

The reaction stoichiometry is 1:1

The amount of NaHCO₃ required is 3.57x10^-3 mol

Molecular weight of NaHCO₃ is 84g/mol.

So 3.57x10^-3mol of NaHCO₃ corresponds to 3.57x10^-3mol x 84g/mol = 0.3g

So the amount of NaHCO₃ required is 0.3g

HCl requirement

1M HCl corresponds to 1 mol/L, so the requirement of 3.57x10^-3mol corrsponds to 3.57x10^-3 L which is equal to 3.57mL

So 3.57 mL of 1M HCl and 0.3g of NaHCO3 is required to produce 80mL of CO₂