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.

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Answer:

Percent of sodium nitrate is 82.13%

Explanation:

[tex]NaNO_{3}\rightarrow NaNO_{2}+\frac{1}{2}O_{2}[/tex]

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

Molar mass of [tex]NaNO_{2}[/tex] = 68.9953 g/mol

Molar mass of [tex]NaNO_{3}[/tex] = 84.9947 g/mol

So, 0.2820 g [tex]NaNO_{2}[/tex] = [tex]\frac{0.2820}{68.9953}[/tex] moles of  [tex]NaNO_{2}[/tex]  = 0.004087 moles of  [tex]NaNO_{2}[/tex]

Above equation suggests, 1 mol of  [tex]NaNO_{2}[/tex]  is formed from 1 mol of  [tex]NaNO_{3}[/tex]

So, 0.004087 moles of  [tex]NaNO_{2}[/tex] are formed from 0.004087 moles of  [tex]NaNO_{3}[/tex]

So, mass of 0.004087 moles of  [tex]NaNO_{3}[/tex] = [tex](0.004087\times 84.9947)g=0.3474g[/tex] [tex]NaNO_{3}[/tex]

% of [tex]NaNO_{3}[/tex]  in sample = [tex]\frac{0.3474}{0.4230}\times 100[/tex]% = [tex]82.13[/tex]%