Question

Calculate the mass of CO2 produced by the following reactionNaHCO3 (s) + HC2H3O2 (aq) → Na+ (aq) + C2H3O2 (aq) + H2O (1) +CO2 (g)if you start with 2.59 g of baking soda NaHCO3=(Molar mass of NaHCO3 = 84.00 g/mol and molar mass of CO2 = 44.01g/mol)

Answer 1

1) Balance the equation.

NaHCO3 + HC2H3O2 => Na+ + C2H3O2 + H2O + CO2

2) Find out the molar mass of NaHCO3

Na = 22.990 g/mol

H = 1.00794 g/mol

C = 12.0107 g/mol

O = 15.9994 g/mol

The molar mass of NaHCO3 is

(22.990) + 1.00974 + 12.0107 + (15.9994 * 3) = 84.0066 g/mol

3) Convert 2.59 g of NaHCO3 into moles

`\text{moles of NaHCO3 = }2.59\text{ g}\cdot\frac{1\text{ mole of NaHCO3 }}{84.0066\text{ g of NaHCO3}}\text{ = }0.03083\text{ moles of NaHCO3}`

4) Use use the stoichiometric relationship between NaHCO3 and CO2 to find out the potential production of CO2

`molesofCO2=0.03083molesofNaHCO3\cdot\frac{1\text{ mole of CO2}}{1\text{ mole of NaHCO3}}=\text{ 0.03083 moles of CO2}`

5) Convert moles of CO2 into grams of CO2. We need the CO2 molar mass.

The molar mass of CO2 is = 44.0095 g/mole

`\text{grams of CO2 = 0.03083 moles of CO2}\cdot\frac{44.01\text{ grams of CO2}}{1\text{ mole of CO2}}\text{ = 1.36 g of CO2}`