1. Consider the reaction: NaHCO3(aq) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l) Question: 0.056 moles...

Question

Question

1. Consider the reaction:
NaHCO₃(aq) + HCl(aq) → NaCl(aq) + CO₂(g) + H₂O(l)

Question: 0.056 moles of NaHCO₃ are reacted with excess HCl. 2.267 g of CO₂ were produced. What is the percent yield of CO₂?

Answer 1

Answer #1

the balanced equation is as follows

NaHCO3 + HCl -----------------------> NaCl + CO2 + H2O

1 mol NaHCO3 ----------------------------------------> 1 mol CO2

0.056 mol ---------------------------------------------->?

=> 0.056 * 1 / 1

=> 0.056 moles

mass = moles * molar mass

theoretical yield = 0.056 mol * 44 g/mol => 2.464 g

percent yield = actual yield / theoretical yield * 100

percent yield = 2.267 / 2.464 *100 => 92 %

answer => 92 %

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Answer 2

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