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Chapter 3: Problem 144

A \(0.4230-\mathrm{g}\) sample of impure sodium nitrate was heated, converting all the sodium nitrate to \(0.2864 \mathrm{~g}\) of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the original sample.

Short Answer

Expert verified
The percent of sodium nitrate in the original sample is 83.40%, calculated by finding the moles of sodium nitrite formed, using the 1:1 stoichiometric ratio to determine the moles of sodium nitrate, calculating the mass of sodium nitrate, and dividing by the mass of the original sample.

Step by step solution

01

1. Calculate the molecular weight of sodium nitrite and sodium nitrate

The molecular weights for the sodium nitrite (NaNO2) and sodium nitrate (NaNO3) are calculated as follows: NaNO2 = 1(22.99 g/mol Na) + 1(14.01 g/mol N) + 2(16.00 g/mol O) = 69.00 g/mol NaNO3 = 1(22.99 g/mol Na) + 1(14.01 g/mol N) + 3(16.00 g/mol O) = 85.00 g/mol
02

2. Calculate the moles of sodium nitrite formed in the reaction

Given the mass of sodium nitrite formed is 0.2864 g, we can now calculate the moles of sodium nitrite using the molecular weight: Moles of NaNO2 = mass / molecular weight Moles of NaNO2 = 0.2864 g / 69.00 g/mol Moles of NaNO2 = 0.00415 mol
03

3. Calculate the moles of sodium nitrate in the original sample

Since the balanced chemical equation indicates a 1:1 ratio between sodium nitrite and sodium nitrate, the moles of sodium nitrate that reacted will be the same as the moles of sodium nitrite formed: Moles of NaNO3 = 0.00415 mol
04

4. Calculate the mass of sodium nitrate in the original sample

Now, we can calculate the mass of sodium nitrate in the original sample using its molecular weight: Mass of NaNO3 = moles * molecular weight Mass of NaNO3 = 0.00415 mol * 85.00 g/mol Mass of NaNO3 = 0.35275 g
05

5. Calculate the percentage of sodium nitrate in the original sample

Finally, we can determine the percentage of sodium nitrate in the original sample using the mass of the original sample (0.4230 g) and the mass of sodium nitrate (0.35275 g) we just calculated: Percentage of NaNO3 = (mass of NaNO3 / mass of original sample) * 100 Percentage of NaNO3 = (0.35275 g / 0.4230 g) * 100 Percentage of NaNO3 = 83.40% The percent of sodium nitrate in the original sample is 83.40%.

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Most popular questions from this chapter

The reusable booster rockets of the U.S. space shuttle employ a mixture of aluminum and ammonium perchlorate for fuel. A possible equation for this reaction is $$ \begin{aligned} 3 \mathrm{Al}(s)+3 \mathrm{NH}_{4} \mathrm{ClO}_{4}(s) & \longrightarrow \\ \mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{AlCl}_{3}(s)+3 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \end{aligned} $$ What mass of \(\mathrm{NH}_{4} \mathrm{ClO}_{4}\) should be used in the fuel mixture for every kilogram of \(\overline{\mathrm{Al}}\) ?

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Consider the following reaction: $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ If a container were to have 10 molecules of \(\mathrm{O}_{2}\) and 10 molecules of \(\mathrm{NH}_{3}\) initially, how many total molecules (reactants plus products) would be present in the container after this reaction goes to completion?

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