The salt case: how can we check the entire calculation chain?

Support map

Model diagram

Use only paper, a calculator and supplied data. Do not carry out reactions, generate gases, heat substances or mix household chemicals.

Situation and question

An invented report describes a 13.25 g sample containing 80% Na₂CO₃ by mass; the remainder is inert in this model. It receives 0.150 mol of HCl. The reaction is Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. The dry mixture obtained weighs 8.00 g and contains 87.75% NaCl. Can we calculate salt yield? No actual reaction is required.

New terms, fully explained

Separate three fractions: starting-sample purity, NaCl content of the collected mixture and percentage yield. Calculate reacting amount from the reacting substance’s own mass. Both reactants constrain theoretical product. Correct collected-mixture mass for its composition to find actual target-product mass.

Predict first

May all 13.25 g be treated as Na₂CO₃ and all 8.00 g as NaCl? Is 0.150 mol of HCl enough to convert 0.100 mol of carbonate completely? State the required ratio before calculating any salt mass.

Investigate or analyse given data

Use M(Na₂CO₃)=106 and M(NaCl)=58.5 g/mol. The data specify 0.150 mol of HCl substance, not the mass or volume of its solution. Inertness is a supplied model assumption, not a claim about every unidentified impurity. Composition is given from separate analysis.

Worked example

Na₂CO₃ mass is 13.25×0.80=10.60 g, or 0.100 mol. Compare 0.100/1 with 0.150/2=0.0750 mol: HCl limits. Theoretical n(NaCl)=2×0.0750=0.150 mol and m=8.775 g. Actual NaCl is 8.00×0.8775=7.020 g. Yield is 7.020/8.775×100%=80.0%.

A common mistake

Mistake: calculating yield as 8.00/8.775 or using all the carbonate as the limit. The ideal model leaves 0.100−0.0750=0.0250 mol, or 2.65 g, of Na₂CO₃ in addition to the original inert impurity. These leftovers are not pure salt mass. Yield alone cannot establish a unique cause of losses.

Transfer to a new case

Transfer: the starting sample is 6.625 g with the same 80% Na₂CO₃, and HCl is now 0.200 mol. The collected mixture is 5.00 g containing 93.6% NaCl. Find the limiting reactant, theoretical and actual NaCl masses and percentage yield. Also calculate the remaining HCl amount.

Answer and recall

Check: Na₂CO₃ is 5.30 g = 0.0500 mol and limits; theoretical NaCl is 0.100 mol = 5.85 g. Actual mass is 5.00×0.936=4.68 g, giving 80% yield. The ideal model leaves 0.100 mol of HCl. Hide the answers and reconstruct each step with units; the next block moves to water and solutions.

Exercise

Why does HCl limit the main case? Is 8.00 g pure salt? What does 80% mean?

Answers

When to revisit

Close the page and, after 10 minutes, reconstruct the map, define the terms in your own words and solve the new case. Repeat without hints after a day and a week. If an error persists, return to the worked example and write down the missing step.

Sources and limits

• OpenStax: reaction stoichiometry
• OpenStax: limiting reactants and yields

Why can the collected mass differ from the calculated mass?