Question

nitrogen and oxygen gases react to form dinitrogen oxide gas (n₂o). what volume of o₂ is needed to produce 7.84 l of n₂o?
balanced equation:
□ + □ → □
part 1 of 2

Explanation:

Step1: Balance the chemical equation

The reaction is between nitrogen ($N_2$) and oxygen ($O_2$) to form dinitrogen oxide ($N_2O$). Let's balance the equation.

First, write the unbalanced equation: $N_2 + O_2
ightarrow N_2O$

To balance oxygen, we can see that on the right side, we have 1 O in $N_2O$, and on the left side, we have 2 O in $O_2$. So we need to adjust the coefficients. Let's multiply $N_2O$ by 2 to get 2 O atoms on the right: $N_2 + O_2
ightarrow 2N_2O$

Now, balance nitrogen. On the right, we have 4 N (from $2N_2O$), so on the left, we need 2 $N_2$ molecules: $2N_2 + O_2
ightarrow 2N_2O$

Wait, let's check oxygen again. On the left, we have 2 O (from $O_2$), and on the right, we have 2 O (from $2N_2O$). Wait, no, $2N_2O$ has 2 O atoms? No, each $N_2O$ has 1 O, so 2 $N_2O$ has 2 O. And $O_2$ has 2 O. So that's balanced? Wait, no, let's do it properly.

The correct balanced equation for the formation of $N_2O$ from $N_2$ and $O_2$ is:

First, the reaction: $N_2 + O_2
ightarrow N_2O$

To balance O: multiply $N_2O$ by 2: $N_2 + O_2
ightarrow 2N_2O$

Now, N: on the right, 4 N, so on the left, we need 2 $N_2$: $2N_2 + O_2
ightarrow 2N_2O$

Now, O: on the left, 2 O (from $O_2$), on the right, 2 O (from $2N_2O$). Wait, that's balanced? Wait, no, $2N_2O$ has 2 O atoms? Each $N_2O$ is N₂O, so 2 N₂O is 2*1 = 2 O. And O₂ is 2 O. So yes, that's balanced. Wait, but actually, the correct balanced equation is $2N_2 + O_2 = 2N_2O$? Wait, no, let's check the oxidation states. N in $N_2$ is 0, in $N_2O$ is +1. O in $O_2$ is 0, in $N_2O$ is -2. So the number of electrons transferred: each N atom loses 1 electron (from 0 to +1), so 2 N atoms (in $N_2$) lose 2 electrons. O in $O_2$ gains 2 electrons per O atom (from 0 to -2), so 2 O atoms (in $O_2$) gain 4 electrons. Wait, that's a problem. So maybe the correct balanced equation is different. Wait, maybe I made a mistake. Let's look up the correct balanced equation for the formation of dinitrogen oxide from nitrogen and oxygen.

The correct balanced chemical equation for the reaction of nitrogen gas ($N_2$) and oxygen gas ($O_2$) to form dinitrogen oxide ($N_2O$) is:

$2N_2 + O_2
ightarrow 2N_2O$

Wait, but let's check the number of atoms:

Left side: 2 $N_2$ (4 N) and 1 $O_2$ (2 O)

Right side: 2 $N_2O$ (4 N and 2 O)

Yes, that's balanced. So the balanced equation is $2N_2 + O_2
ightarrow 2N_2O$

Step2: Determine the mole ratio (volume ratio for gases)

For gases at the same temperature and pressure, the volume ratio is equal to the mole ratio (Avogadro's law). So from the balanced equation:

$2N_2 + O_2
ightarrow 2N_2O$

The mole ratio of $O_2$ to $N_2O$ is 1 : 2. That is, 1 mole of $O_2$ produces 2 moles of $N_2O$, so the volume ratio of $O_2$ to $N_2O$ is also 1 : 2 (since volume is proportional to moles for gases under same conditions).

Step3: Calculate the volume of $O_2$ needed

We need to produce 7.84 L of $N_2O$. Let $V_{O_2}$ be the volume of $O_2$ needed.

From the volume ratio: $\frac{V_{O_2}}{V_{N_2O}} = \frac{1}{2}$

So $V_{O_2} = \frac{1}{2} \times V_{N_2O}$

Substitute $V_{N_2O} = 7.84$ L:

$V_{O_2} = \frac{1}{2} \times 7.84$ L = 3.92 L

Answer:

The volume of $O_2$ needed is $\boxed{3.92}$ L.

(Note: First, we balanced the equation as $2N_2 + O_2
ightarrow 2N_2O$, then used the volume ratio from the balanced equation (1 mole $O_2$ produces 2 moles $N_2O$, so volume ratio 1:2) to calculate the volume of $O_2$ needed for 7.84 L of $N_2O$.)