# 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?

Jan 29, 2016

$\text{1.31 g CO"_2}$ will be produced.

#### Explanation:

Balanced Equation

$\text{3NaHCO"_3+"H"_3"C"_6"H"_5"O"_7}$$\rightarrow$$\text{Na"_3"C"_6"H"_5"O"_7+"3CO"_2+"3H"_2"O}$

We will need the molar masses of $\text{NaHCO"_3}$ and $\text{CO"_2}$, and their mole ratio as well.

Molar Masses

$\text{NaHCO"_3} :$$\text{84.006609 g/mol}$

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

$\text{CO"_2} :$$\text{44.0095 g/mol}$

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

Mole Ratio

From the balanced equation, the mole ratio between
$\text{NaHCO"_3}$ and $\text{CO"_2}$ is ${\text{3 mol NaHCO}}_{3} :$$\text{3 mol CO"_2}$.

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

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