Question

Exactly 0.2220 g of pure Na2CO3 was dissolved in 100.0 mL of 0.0731 M HCl. (a) What mass in grams of CO2 was evolved

Answer 1

This problem is providing information about a chemical reaction between sodium carbonate and hydrochloric acid. Thus, the mass of evolved carbon dioxide is required and found to be 0.0922 g according to:

Stoichiometry:

In chemistry, stoichiometry is a tool that allows us to perform mole-mass calculations based on proportional factors involving the balanced chemical equation, molar masses and mole ratios. In such a way, for the reaction in this problem, we write:

$2HCl+Na_2CO_3\rightarrow 2NaCl+H_2O+CO_2$

In such a way, since both HCl and Na2CO3 are given, we calculate the grams of CO2 produced by each one, to identify the limiting reactant, according to the 2:1 and 1:1 mole ratios the reactants have with CO2 respectively:

$0.2220gNa_2CO_3*\frac{1molNa_2CO_3}{106gNa_2CO_3} *\frac{1molCO_2}{1molNa_2CO_3}*\frac{44.01 gCO_2}{1molCO_2}= 0.0922gCO_2\\ \\ 0.1000L*0.0731\frac{molHCl}{L} *\frac{1molCO_2}{2molHCl}*\frac{44.01 gCO_2}{1molCO_2}=0.161gCO_2$

Thus, since Na2CO3 produce the smallest amount of CO2, we infer it is the limiting reactant and therefore the correct produced amount is 0.0922 g.

Learn more about stoichiometry: brainly.com/question/9743981