$2 V O + 3 F e_{2} O_{3} \to 6 F e O + V_{2} O_{5}$ .
If we start with 2 g of $V O$ and 6 g of $F e_{2} O_{3}$ , is the limiting reagent. [V = 48 g] [Fe = 56 g]

V O

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F e_{2} O_{3}

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F e_{2} O

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V_{2} O_{5}

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Solution

The correct option is B $F e_{2} O_{3}$
If you check, the reaction is already balanced.
2 moles of $V O$ requires 3 moles of $F e_{2} O_{3}$
The ratio of moles of $V O$ to $F e_{2} O_{3}$ is $\frac{2}{3}$ . If the actual ratio is lower than this, that means that there is a lower amount of $V O$ , making it the limiting reagent. If this is higher than 0.67, then $F e_{2} O_{3}$ is the limiting reagent. Let's calculate!
Moles of $V O$ = $\frac{2 g}{(48 + 16) g / m o l} = \frac{1}{32} m o l$
Moles of $F e_{2} O_{3}$ = $\frac{6 g}{(112 + 48) g / m o l} = \frac{3}{80} m o l$
Ratio of $V O$ to $F e_{2} O_{3}$ = 0.83
The ratio is higher than 0.67, making $F e_{2} O_{3}$ the limiting reagent.

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$2 V O + 3 F e_{2} O_{3} \to 6 F e O + V_{2} O_{5}$ .
If we start with 2 g of $V O$ and 6 g of $F e_{2} O_{3}$ , is the limiting reagent. [V = 48 g] [Fe = 56 g]

$2 V O + 3 F e_{2} O_{3} \to 6 F e O + V_{2} O_{5}$ If we start with 2g of VO and 5.75g of $F e_{2} O_{3}$ , which is the limiting reagent?

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$2 V O + 3 F e_{2} O_{3} \to 6 F e O + V_{2} O_{5}$ .
If we start with 2 g of $V O$ and 6 g of $F e_{2} O_{3}$ , is the limiting reagent. [V = 48 g] [Fe = 56 g]