# A reaction produces 0.813 moles of H_2O. How many molecules are produced?

Aug 14, 2016

$0.813 \times {N}_{A}$, where ${N}_{A} = \text{Avogadro's number}$.

#### Explanation:

And ${N}_{A} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.

Why should we use such an absurdly large number? Well, it happens that $\text{Avogadro's number}$ of ""^1H atoms have a mass of $1.00 \cdot g$ precisely. ${N}_{A}$ is thus the link between the macro world of grams and kilograms and litres, that which we can measure in a laboratory by some means, to the micro world of atoms and molecules, that which we can only infer; and its use allows us to measure equivalent masses of elements and compounds practically, where given massess specify precise numbers of atoms. If you can digest this, you will have mastered most of A level chemistry. Please test your understanding.

${H}_{2} + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow {H}_{2} O \left(l\right)$

Given the equation, if we get $18 \cdot g$ of water product, i.e. $1$ $m o l$ of water molecules, can you determine (i) the mass, and (ii) the number of constituent atoms of the reactants?