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.
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
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
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