I need to calculate the mass of magnesium needed to produce $\pu{5 kg}$ of vanadium from the following reaction:

$$\ce{3Mg + 2VCl3 -> 2V + 3MgCl2}$$

given relative atomic masses for $\ce{Mg}$ and $\ce{V}$ as $24$ and $51$, respectively.

I multiplied $24 × 3$ and $51 × 2$ to get

$72 : 102 $

I changed $\pu{5 kg}$ into grams, which is $5000$.

I cross-multiplied to find the answer and then divided the answer by $1000$ to turn it back to kilograms because the question stated it. My answer is $\pu{3.5 kg}$. Is this correct?


You answer is correct.

It is given that for every 3 moles of $\ce{Mg}$ that reacts, 2 moles of $\ce{V}$ are produced. Here's one possible way to proceed towards the answer.

First, find out how many moles of $\ce{V}$ are in a mass of $5000\,\mathrm{g}$:

$$5000\,\mathrm{g}\cdot\left({1\,\mathrm{mol}\,\ce{V}}\over 51\,\mathrm{g}\right) = 98\,\mathrm{mol}\,\ce{V}$$

Second, calculate how many moles of $\ce{Mg}$ will yield $98$ moles of $\ce{V}$ by multiplying the ratio of the stoichiometric coefficients (3:2) by the number from step 1:

$$\left({3\,\mathrm{mol}\,\ce{Mg}\over 2\,\mathrm{mol}\,\ce{V}}\right)\cdot 98\,\mathrm{mol}\,\ce{V} = 147\,\mathrm{mol}\,\ce{Mg}$$

Finally, calculate the mass of $\ce{Mg}$ by multiplying the number of moles from step 2 by the atomic mass:

$$147\,\mathrm{mol}\,\ce{Mg}\cdot\left({24\,\mathrm{g}\over 1\,\mathrm{mol}\,\ce{Mg}}\right) = 3528\,\mathrm{g} \approx 3.5\,\mathrm{kg}$$


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