Question

How many molecules (not moles) of NH3 are produced from 5.01×10−4 g of H2?

Answer 1

The chemical formula of ammonia is written as NH3 we can just use the ratio of the atoms since we are not given the reaction. We also make use the avogadro's number to convert from moles to molecules. We calculate as follows:

5.01 x 10^-4 g H2 ( 1 mol / 2.02 g H2 ) ( 2 mol NH3 / 3 mol H2 ) ( 6.022x10
^23 molecules / 1 mol ) = 9.96 x 10^19 molecules NH3

Answer 2

Answer : The number of molecules of $NH_3$ produced are $10.06\times 10^{19}$

Explanation :

The given balanced chemical reaction will be,

$N_2+3H_2\rightarrow 2NH_3$

First we have to calculate the moles of $H_2$.

$\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{5.01\times 10^{-4}g}{2g/mole}=2.50\times 10^{-4}moles$

Now we have to calculate the moles of $NH_3$.

From the balanced chemical reaction we conclude that,

As, 3 moles of $H_2$ react to produce 2 moles of $NH_3$

So, $2.50\times 10^{-4}moles$ of $H_2$ react to produce $\frac{2}{3}\times 2.50\times 10^{-4}=1.67\times 10^{-4}moles$ moles of $NH_3$

Now we have to calculate the number of molecules of $NH_3$.

As, 1 mole of $NH_3$ produces $6.022\times 10^{23}$ molecule of $NH_3$

So, $1.67\times 10^{-4}moles$ of $NH_3$ produces $(1.67\times 10^{-4})\times (6.022\times 10^{23})=10.06\times 10^{19}$ molecule of $NH_3$

Therefore, the number of molecules of $NH_3$ produced are $10.06\times 10^{19}$