Question

How many grams of carbon dioxide are produced when 2.50 g of sodium hydrogen carbonate react with excess citric acid according to the equation `3NaHCO_3+H_3C_6H_5O_7->Na_3C_6H_5O_7 + 3CO_2 + 3H_2O`?

Answer 1

`"1.31 g CO"_2"` will be produced.

Explanation:

Balanced Equation

`"3NaHCO"_3+"H"_3"C"_6"H"_5"O"_7"` `rarr` `"Na"_3"C"_6"H"_5"O"_7+"3CO"_2+"3H"_2"O"`

We will need the molar masses of `"NaHCO"_3"` and `"CO"_2"`, and their mole ratio as well.

Molar Masses

`"NaHCO"_3":` `"84.006609 g/mol"`

http://pubchem.ncbi.nlm.nih.gov/compound/516892#section=Top

`"CO"_2":` `"44.0095 g/mol"`

http://pubchem.ncbi.nlm.nih.gov/compound/280#section=Top

Mole Ratio

From the balanced equation, the mole ratio between
`"NaHCO"_3"` and `"CO"_2"` is `"3 mol NaHCO"_3:` `"3 mol CO"_2"`.

Stoichiometric Equation
Determine the moles of `"NaHCO"_3"` by dividing the given mass of `"NaHCO"_3"` by its molar mass, then determine moles of `"CO"_2"` by multiplying times the mole ratio with `"CO"_2"` in the numerator, then multiply times the molar mass of `"CO"_2"`.

`2.50cancel"g NaHCO"_3xx(1cancel"mol NaHCO"_3)/(84.006609cancel"g NaHCO"_3)xx(cancel3^1cancel"mol CO"_2)/(cancel3^1cancel"mol NaHCO"_3)xx(44.0095"g CO"_2)/(1cancel"mol CO"_2)="1.31 g CO"_2"` rounded to three significant figures