Question

a compound contains 24.9g of iron (fe) and it weighs 80.0g. assuming that the each one mole of the molecule contains one mole of iron, what is its molecular weight, in g/mol?
17.4g/mol
0.00557g/mol
179g/mol
0.0575g/mol
multiple choice 10 points
dinitrogen pentoxide (n₂o₅) decomposes according to: 2 n₂o₅ → 4 no₂ + o₂
how many grams of no₂ are made when 2.03g of oxygen (o₂) are produced?
0.353g
11.7g
0.730g
5.65g

Explanation:

Step1: Calculate moles of iron

The molar mass of iron (Fe) is approximately 55.85 g/mol. The number of moles of iron $n_{Fe}=\frac{m_{Fe}}{M_{Fe}}$, where $m_{Fe} = 24.9g$ and $M_{Fe}=55.85g/mol$. So $n_{Fe}=\frac{24.9g}{55.85g/mol}\approx0.446mol$.

Step2: Determine moles of compound

Since each mole of the molecule contains one - mole of iron, the number of moles of the compound $n$ is equal to the number of moles of iron, $n = 0.446mol$.

Step3: Calculate molecular weight of compound

The mass of the compound $m = 80.0g$. The molecular weight (molar mass) $M=\frac{m}{n}$. Substituting $m = 80.0g$ and $n = 0.446mol$, we get $M=\frac{80.0g}{0.446mol}\approx179g/mol$.

Step4: For the second part - Calculate moles of oxygen

The molar mass of $O_2$ is $M_{O_2}=32g/mol$. The number of moles of $O_2$ produced, $n_{O_2}=\frac{m_{O_2}}{M_{O_2}}$, where $m_{O_2} = 2.03g$. So $n_{O_2}=\frac{2.03g}{32g/mol}\approx0.0634mol$.

Step5: Determine moles of $NO_2$ based on mole - ratio

From the balanced chemical equation $2N_2O_5
ightarrow4NO_2 + O_2$, the mole - ratio of $NO_2$ to $O_2$ is $\frac{n_{NO_2}}{n_{O_2}}=\frac{4}{1}$. So $n_{NO_2}=4\times n_{O_2}=4\times0.0634mol = 0.2536mol$.

Step6: Calculate mass of $NO_2$

The molar mass of $NO_2$ is $M_{NO_2}=46g/mol$. The mass of $NO_2$, $m_{NO_2}=n_{NO_2}\times M_{NO_2}=0.2536mol\times46g/mol\approx11.7g$.

Answer:

1. C. 179g/mol
2. B. 11.7g