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
${\text{3NaHC}}{{\text{O}}_{\text{3}}}\,{\text{ + }}\,{{\text{H}}_{\text{3}}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{{\text{O}}_{\text{7}}} \to \,{\text{N}}{{\text{a}}_{\text{3}}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{{\text{O}}_{\text{7}}}\,{\text{ + }}\,{\text{3C}}{{\text{O}}_{\text{2}}}{\text{ + 3}}{{\text{H}}_{\text{2}}}{\text{O}}$ ?

Answer 1

Determine the yield means we have to determine the amount of the substance. In a chemical reaction, the yield is determined by using the stoichiometry between the reactants and the products.Stoichiometry is nothing but the mole ratio of reactants and products involved in the balanced chemical reaction.Here, the mole ratio is typically the ratio of integer numbers. In many chemical reactions, the stoichiometric relation is used to determine the amount that is the yield of the product.

Formula used: The moles of the substance are calculated using the following formula:
${\text{mole = }}\dfrac{{{\text{weight}}}}{{{\text{molar}}\,{\text{mass}}}}$

Complete step-by-step solution: The molecular formula of the sodium hydrogen carbonate is ${\text{NaHC}}{{\text{O}}_{\text{3}}}$, citric acid is ${{\text{H}}_{\text{3}}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{{\text{O}}_{\text{7}}}$, and the sodium citrate is ${\text{N}}{{\text{a}}_{\text{3}}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{{\text{O}}_{\text{7}}}$, carbon dioxide is ${\text{C}}{{\text{O}}_{\text{2}}}$, and water is ${\text{3}}{{\text{H}}_{\text{2}}}{\text{O}}$.
Here, we can see that reaction given is a balanced chemical reaction which is as follows:
${\text{3NaHC}}{{\text{O}}_{\text{3}}}\,{\text{ + }}\,{{\text{H}}_{\text{3}}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{{\text{O}}_{\text{7}}} \to \,{\text{N}}{{\text{a}}_{\text{3}}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{{\text{O}}_{\text{7}}}\,{\text{ + }}\,{\text{3C}}{{\text{O}}_{\text{2}}}{\text{ + 3}}{{\text{H}}_{\text{2}}}{\text{O}}$
Here, excess of the citric acid is given hence to determine the amount of the carbon dioxide produced use amount of the sodium hydrogen carbonate which acts as limiting reagent.
The mole ratio between the sodium hydrogen carbonate and the carbon dioxide is as follows:
${\text{3}}\,{\text{mol}}\,\,\,{\text{NaHC}}{{\text{O}}_{\text{3}}}\,:\,{\text{3}}\,{\text{mol}}\,\,{\text{C}}{{\text{O}}_{\text{2}}}$
It indicates the same moles of carbon dioxide are produced from the same moles of sodium hydrogen carbonate.
${\text{1}}\,{\text{mol}}\,\,\,{\text{NaHC}}{{\text{O}}_{\text{3}}}\,:\,{\text{1}}\,{\text{mol}}\,\,{\text{C}}{{\text{O}}_{\text{2}}}$
The molar mass of the sodium ${\text{NaHC}}{{\text{O}}_{\text{3}}}$ is $84{\text{gmo}}{{\text{l}}^{ - 1}}$, and ${\text{C}}{{\text{O}}_{\text{2}}}$ is $44{\text{gmo}}{{\text{l}}^{ - 1}}$.
${\text{1}}\,{\text{mol}}\,\,\,{\text{NaHC}}{{\text{O}}_{\text{3}}}\, = \,{\text{1}}\,{\text{mol}}\,\,{\text{C}}{{\text{O}}_{\text{2}}}$
$84\,{\text{gram}}\,\,\,{\text{NaHC}}{{\text{O}}_{\text{3}}}\, = \,44\,{\text{gram}}\,\,{\text{C}}{{\text{O}}_{\text{2}}}$
$2.50\,{\text{gram}}\,\,\,{\text{NaHC}}{{\text{O}}_{\text{3}}}\, = \,\dfrac{{2.50\,{\text{gram}}\,\,\,{\text{NaHC}}{{\text{O}}_{\text{3}}} \times 44\,{\text{gram}}\,\,{\text{C}}{{\text{O}}_{\text{2}}}}}{{84\,{\text{gram}}\,\,\,{\text{NaHC}}{{\text{O}}_{\text{3}}}}}$
$2.50\,{\text{gram}}\,\,\,{\text{NaHC}}{{\text{O}}_{\text{3}}}\, = \,1.3095\,{\text{gram}}\,\,{\text{C}}{{\text{O}}_{\text{2}}}$
Thus, grams of carbon dioxide produced are $1.31\,{\text{gram}}\,$.

Note: The problem is based on the limiting reagent. The limiting reagent is the reagent in the reaction which determines the amount of the product formed or whose amount limits the product formation.The amount of the product is determined using the mole ratio between the limiting reagent and the product.The excess reagent is the reagent that is present in an excess amount which is not used to determine the amount of the products.