Question

How many moles of $H_{2}O$ are formed by $10$ moles of $HN{O}_3$ and $10$ moles $Cu$ in the following reaction?
$3Cu+8HN{O}_3⟶3Cu(N{O}_3{)}_2+2NO+4H_{2}O$

Answer 1

In this question, first, we have to find the limiting reagent.

Limiting Reagent:

-> Available in less than the stoichiometric amount.

-> Gets completely consumed in the reaction

-> Decides the amount of product.

So, according to the given reaction:

$3Cu+8HN{O}_3\to 3Cu(N{O}_3{)}_2+2NO+4H_{2}O$

We are given in question, $10$ moles each of $HN{O}_3$ and $Cu$

We know that $3$ moles of $Cu$ require $8$ moles of $HN{O}_3$

So, $10$ moles of $Cu$ will require $=\frac{10\times 8}{3}$ moles of $HN{O}_3$

$=26.66$ moles of $HN{O}_3$

Similarly, $8$ moles of $HN{O}_3$ require $3$ moles of $Cu$

So, $10$ moles of $HN{O}_3$ will require $=\frac{10\times 3}{8}$ moles of $Cu$

$3.75$ moles of $Cu$

Therefore, since $10$ moles of $Cu$ will require $26.66$ moles of $HN{O}_3$ and we only have $10$ moles of $HN{O}_3$. So, $HN{O}_3$ is the limiting reagent anf the reaction will take place accordingly.

Now, since $8$ moles of $HN{O}_3$ form $4$ moles of $H_{2}O$

So, $10$ moles of $HN{O}_3$ will form $=\frac{10\times 4}{8}$ moles of $H_{2}O$

$=5$ moles of $H_{2}O$