practice problems
1. nitrogen can combine with oxygen to form several different oxides of nitrogen. these oxides include no₂, no, n₂o.
a) how many moles of o₂ are required to react with 9.35×10⁻² moles of n₂ to form n₂o?
b) how many moles of o₂ are required to react with 9.35×10⁻² moles of n₂ to form no₂?
2. 2c₂h₆ + 7o₂ → 4co₂ + 6h₂o
a) how many moles of o₂ are required to react with 13.9 mol of c₂h₆? (ans. 48.7 mol)
b) how many moles of water would be produced by 1.40 mol of o₂? (ans. 1.2 mol)

Question

practice problems
1. nitrogen can combine with oxygen to form several different oxides of nitrogen. these oxides include no₂, no, n₂o.
   a) how many moles of o₂ are required to react with 9.35×10⁻² moles of n₂ to form n₂o?
   b) how many moles of o₂ are required to react with 9.35×10⁻² moles of n₂ to form no₂?
2. 2c₂h₆ + 7o₂ → 4co₂ + 6h₂o
   a) how many moles of o₂ are required to react with 13.9 mol of c₂h₆? (ans. 48.7 mol)
   b) how many moles of water would be produced by 1.40 mol of o₂? (ans. 1.2 mol)

Accepted Answer

### Problem 1a:
# Explanation:
## Step1: Write the balanced equation for $N_2$ and $O_2$ forming $N_2O$
The balanced equation is $2N_2 + O_2
ightarrow 2N_2O$. From the equation, the mole ratio of $N_2$ to $O_2$ is $2:1$.
## Step2: Use the mole ratio to calculate moles of $O_2$
Given moles of $N_2 = 9.35\times 10^{-2}$ mol. Let moles of $O_2$ be $x$. Using the ratio $\frac{n_{N_2}}{n_{O_2}}=\frac{2}{1}$, we have $n_{O_2}=\frac{n_{N_2}}{2}$.
Substitute $n_{N_2} = 9.35\times 10^{-2}$ mol: $n_{O_2}=\frac{9.35\times 10^{-2}}{2}= 4.675\times 10^{-2}$ mol.
# Answer: $4.675\times 10^{-2}$ mol

### Problem 1b:
# Explanation:
## Step1: Write the balanced equation for $N_2$ and $O_2$ forming $NO_2$
The balanced equation is $N_2 + 2O_2
ightarrow 2NO_2$. The mole ratio of $N_2$ to $O_2$ is $1:2$.
## Step2: Use the mole ratio to calculate moles of $O_2$
Given moles of $N_2 = 9.35\times 10^{-2}$ mol. Let moles of $O_2$ be $x$. Using the ratio $\frac{n_{N_2}}{n_{O_2}}=\frac{1}{2}$, we get $n_{O_2}=2\times n_{N_2}$.
Substitute $n_{N_2} = 9.35\times 10^{-2}$ mol: $n_{O_2}=2\times9.35\times 10^{-2}= 0.187$ mol.
# Answer: $0.187$ mol

### Problem 2a:
# Explanation:
## Step1: Identify the mole ratio from the balanced equation
The balanced equation is $2C_2H_6 + 7O_2
ightarrow 4CO_2 + 6H_2O$. The mole ratio of $C_2H_6$ to $O_2$ is $2:7$.
## Step2: Calculate moles of $O_2$
Given moles of $C_2H_6 = 13.9$ mol. Let moles of $O_2$ be $x$. Using the ratio $\frac{n_{C_2H_6}}{n_{O_2}}=\frac{2}{7}$, we have $n_{O_2}=\frac{7\times n_{C_2H_6}}{2}$.
Substitute $n_{C_2H_6} = 13.9$ mol: $n_{O_2}=\frac{7\times13.9}{2}=\frac{97.3}{2} = 48.65\approx48.7$ mol.
# Answer: $48.7$ mol

### Problem 2b:
# Explanation:
## Step1: Identify the mole ratio from the balanced equation
From $2C_2H_6 + 7O_2
ightarrow 4CO_2 + 6H_2O$, the mole ratio of $O_2$ to $H_2O$ is $7:6$.
## Step2: Calculate moles of $H_2O$
Given moles of $O_2 = 1.40$ mol. Let moles of $H_2O$ be $x$. Using the ratio $\frac{n_{O_2}}{n_{H_2O}}=\frac{7}{6}$, we have $n_{H_2O}=\frac{6\times n_{O_2}}{7}$.
Substitute $n_{O_2} = 1.40$ mol: $n_{H_2O}=\frac{6\times1.40}{7}=\frac{8.4}{7}=1.2$ mol.
# Answer: $1.2$ mol

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QUESTION IMAGE

Question

Explanation:

Problem 1a:

Step1: Write the balanced equation for \(N_2\) and \(O_2\) forming \(N_2O\)

Step2: Use the mole ratio to calculate moles of \(O_2\)

Step1: Write the balanced equation for \(N_2\) and \(O_2\) forming \(NO_2\)

Step2: Use the mole ratio to calculate moles of \(O_2\)

Step1: Identify the mole ratio from the balanced equation

Step2: Calculate moles of \(O_2\)

Answer:

Problem 1b: