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A gummy bear with 2.67 g sucrose (342.3 g/mol) reacts with 7.19 g KClO3 (122.5 g/mol) as C12H22O11 + 8KClO3. Which reactant limits?
ASucrose, since only 0.00780 mol is present
BKClO3, since only 0.0587 mol is present
CSucrose, since its molar mass is the larger
DNeither, since both are in exact proportion
Answer & Solution
Correct answer: B. KClO3, since only 0.0587 mol is present
1. Moles of sucrose equal 2.67 divided by 342.3, which is 0.00780 mol.
2. Moles of KClO3 equal 7.19 divided by 122.5, which is 0.0587 mol.
3. The equation needs 8 mol of KClO3 for every 1 mol of sucrose.
4. So 0.0587 mol of KClO3 can react with 0.0587 divided by 8, that is 0.00734 mol of sucrose.
5. Available sucrose is 0.00780 mol, which is more than the 0.00734 mol needed.
6. Sucrose is therefore in excess and KClO3 is the limiting reactant.
7. Comparing raw masses or molar masses instead of mole ratios is the trap in the other options.
_Source: OpenStax Chemistry (CC BY 4.0), Ch 5 "Thermochemistry", section 5.3 Enthalpy_
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