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For 2Al + 3I2 giving 2AlI3, how many moles of I2 react with 0.429 mol of Al?
A0.286 mol
B0.644 mol
C0.429 mol
D1.288 mol
Answer & Solution
Correct answer: B. 0.644 mol
1. The balanced equation gives 3 mol I2 for every 2 mol Al.
2. The stoichiometric factor is therefore 3 mol I2 per 2 mol Al.
3. Multiply the given amount: 0.429 mol Al times 3 divided by 2.
4. That is 1.287 divided by 2, which equals 0.6435 mol.
5. Rounded to three significant figures the answer is 0.644 mol I2.
6. Using the factor upside down would give 0.286 mol, which is the trap here.
_Source: OpenStax Chemistry (CC BY 4.0), Ch 4 "Stoichiometry of Chemical Reactions", section 4.3 Reaction Stoichiometry_
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