Question #195844

If 5 moles of N_{2 }reacted with 12 moles of O_{2 }according to 0 N2 + 2 O_{2 }+ NO_{2 }how many moles NO_{2 }are produced

Expert's answer

**Solution:**

Balanced chemical equation:

N_{2} + 2O_{2 }→ 2NO_{2}

According to the equation above: 2×n(N_{2}) = n(O_{2}) = n(NO_{2})

Moles of each reactant:

n(N_{2}) = 5 mol

n(O_{2}) = 12 mol

Choose one reactant and determine how many moles of the other reactant are necessary to completely react with it. Let's choose N_{2}:

n(O_{2}) = 2 × n(N_{2}) = 2 × 5 mol = 10 mol

The calculation above means that we need 10 mol of O_{2} to completely react with N_{2}.

We have 12 mol of O_{2} and therefore more than enough oxygen gas.

Thus oxygen gas (O_{2}) is in excess and nitrogen gas (N_{2}) must be the limiting reagent.

Therefore,

n(NO_{2}) = 2 × n(N_{2}) = 2 × 5 mol = 10 mol

Moles of NO_{2} = 10 mol

**Answer: 10 mol of NO**_{2}** are produced.**

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