Respuesta :
Answer : The percent of sodium nitrate in the original sample is, 31.35 %
Explanation : Given,
Mass of impure [tex]NaNO_3[/tex] = 0.4230 g
Mass of [tex]NaNO_2[/tex] = 0.1080 g
Molar mass of [tex]NaNO_3[/tex] = 84.99 g/mole
Molar mass of [tex]NaNO_2[/tex] = 68.99 g/mole
First we have to calculate the moles of [tex]NaNO_2[/tex].
[tex]\text{Moles of }NaNO_2=\frac{\text{Mass of }NaNO_2}{\text{Molar mass of }NaNO_2}=\frac{0.1080g}{68.99g/mole}=0.00156moles[/tex]
Now we have to calculate the moles of [tex]NaNO_3[/tex].
The balanced chemical reaction is,
[tex]NaNO_3\rightarrow NaNO_2+\frac{1}{2}O_2[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]NaNO_2[/tex] react to give 1 mole of [tex]NaNO_3[/tex]
As, 0.00156 mole of [tex]NaNO_2[/tex] react to give 0.00156 mole of [tex]NaNO_3[/tex]
Now we have to calculate the mass of [tex]NaNO_3[/tex].
[tex]\text{Mass of }NaNO_3=\text{Moles of }NaNO_3\times \text{Molar mass of }NaNO_3[/tex]
[tex]\text{Mass of }NaNO_3=(0.00156mole)\times (84.99g/mole)=0.1326g[/tex]
Now we have to calculate the percent of sodium nitrate in the original sample.
[tex]\% \text{ of }NaNO_3=\frac{\text{Mass of }NaNO_3}{\text{Mass of impure }NaNO_3}\times 100[/tex]
[tex]\% \text{ of }NaNO_3=\frac{0.1326g}{0.4230g}\times 100=31.35\%[/tex]
Therefore, the percent of sodium nitrate in the original sample is, 31.35 %
