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#?

1 Answer
Jan 29, 2016

Answer:

#"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