For the reaction: $7 A + 13 B + 15 C \to 17 P$
If 15 moles of A, 26 moles of B and 30.5 moles of C are taken initially, then the limiting reagent is:

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None of these

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Solution

The correct option is B B
As per stoichiometry, 7 mol of A will react with 13 mol of B and 15 mol of C
Finding the limiting reagent,
$For A = \frac{given moles}{stoichiometric coefficient} = \frac{15}{7} = 2.14$
$For B = \frac{given moles}{stoichiometric coefficient} = \frac{26}{13} = 2.00$
$For C = \frac{given moles}{stoichiometric coefficient} = \frac{30.5}{15} = 2.03$
Since the ratio is the least for B, it will be the limiting reagent.

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For the reaction: 7A+13B+15C→17P If 15 moles of A, 26 moles of B and 30.5 moles of C are taken initially, then the limiting reagent is:

For the reaction: $7 A + 13 B + 15 C \to 17 P$
If 15 moles of A, 26 moles of B and 30.5 moles of C are taken initially, then the limiting reagent is: