What volume (in mL) of a 0.150 M HNO3 solution will completely react with 35.7 mL of a 0.108 M Na2CO3 solution according to the following balanced chemical equation?

Question

Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)

A. Express the volume in milliliters to three significant figures.
the volume of HNO3 solution = 51.4mL

B. In the reaction in Part A, what mass (in grams) of carbon dioxide forms?

Answer 1

Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)
millimoles Na2CO3 = mL x M = 35.7 x 0.108 = 3.856
millimoles HNO3 needed = 3.855 x (2 mol HNO3/1 mol Na2CO3) = 7.711
7.711 mmoles HNO3 = mL x M so
mL = 7.711/0.150 = 51.408 = 51.4 mL

millimoles CO2 = 3.856 mmol Na2CO3 x (1 mol CO2/1 mol Na2CO3) = 3.856 or 0.003856 mols, then grams CO2 = mols CO2 x molar mass CO2 = ? and remember to round to 3 s.f.