Question

A 0.4230 g sample of impure sodium nitrate (contains sodium nitrate plus inert ingredients) was heated, converting all the sodium nitrate to 0.2820 g of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the original sample.

Answer 1

Do you have the answer? hope this is correct.

Answer 2

Answer:

Percent of sodium nitrate is 82.13%

Explanation:

$NaNO_{3}\rightarrow NaNO_{2}+\frac{1}{2}O_{2}$

Number of moles of a compound is the ratio of mass of the compound to molar mass of the compound.

Molar mass of $NaNO_{2}$ = 68.9953 g/mol

Molar mass of $NaNO_{3}$ = 84.9947 g/mol

So, 0.2820 g $NaNO_{2}$ = $\frac{0.2820}{68.9953}$ moles of  $NaNO_{2}$  = 0.004087 moles of  $NaNO_{2}$

Above equation suggests, 1 mol of  $NaNO_{2}$  is formed from 1 mol of  $NaNO_{3}$

So, 0.004087 moles of  $NaNO_{2}$ are formed from 0.004087 moles of  $NaNO_{3}$

So, mass of 0.004087 moles of  $NaNO_{3}$ = $(0.004087\times 84.9947)g=0.3474g$ $NaNO_{3}$

% of $NaNO_{3}$  in sample = $\frac{0.3474}{0.4230}\times 100$% = $82.13$%