Question

consider the combustion of propane: c₃h₈ + 5o₂ → 3co₂ + 4h₂o. if 2.5 moles of c₃h₈ are burned, how many moles of o₂ are consumed?
12.5 moles
25.0 moles
7.5 moles
5.0 moles
how many moles of water (h₂o) are present in 54.0 grams of water? (molar mass of h₂o≈18.0 g/mol)
972 moles
3.0 moles
0.33 moles
54.0 moles
the percentage of nitrogen, by mass, in copper(ii) nitrate, cu(no3)2, is
14.9%
33.9%
36.5%
40.2%

Explanation:

First Sub - Question (Combustion of Propane)

Step 1: Identify the mole ratio

From the balanced equation $\ce{C_{3}H_{8} + 5O_{2}\to 3CO_{2} + 4H_{2}O}$, the mole ratio of $\ce{C_{3}H_{8}}$ to $\ce{O_{2}}$ is $1:5$.

Step 2: Calculate moles of $\ce{O_{2}}$

Given moles of $\ce{C_{3}H_{8}} = 2.5$ moles. Let moles of $\ce{O_{2}}$ be $x$. Using the mole ratio $\frac{\text{moles of } \ce{C_{3}H_{8}}}{\text{moles of } \ce{O_{2}}}=\frac{1}{5}$, we substitute moles of $\ce{C_{3}H_{8}}$: $\frac{2.5}{x}=\frac{1}{5}$. Cross - multiplying gives $x = 2.5\times5=12.5$ moles.

Step 1: Recall the formula for moles

The formula for moles ($n$) is $n=\frac{m}{M}$, where $m$ is mass and $M$ is molar mass.

Step 2: Substitute values

Given $m = 54.0$ g and $M = 18.0$ g/mol. So, $n=\frac{54.0}{18.0}=3.0$ moles.

Step 1: Calculate molar mass of $\ce{Cu(NO_{3})_{2}}$

Molar mass of $\ce{Cu}=63.55$ g/mol, molar mass of $\ce{N}=14.01$ g/mol, molar mass of $\ce{O}=16.00$ g/mol.
For $\ce{Cu(NO_{3})_{2}}$, molar mass $M=63.55 + 2\times(14.01+3\times16.00)=63.55+2\times(14.01 + 48.00)=63.55+2\times62.01=63.55 + 124.02 = 187.57$ g/mol.

Step 2: Calculate mass of nitrogen

There are 2 nitrogen atoms in $\ce{Cu(NO_{3})_{2}}$, so mass of nitrogen $m_N=2\times14.01 = 28.02$ g/mol.

Step 3: Calculate mass percent

Mass percent of nitrogen $=\frac{\text{mass of N in compound}}{\text{molar mass of compound}}\times100=\frac{28.02}{187.57}\times100\approx14.9\%$.

Answer:

12.5 moles (The option is "12.5 moles")

Second Sub - Question (Moles of Water from Mass)