Math / Numbers & Operations

Question $\text{Given the balanced chemical reaction:}$ $\text{N}_2\text{H}_4 + 2\text{H}_2\text{O}_2 \rightarrow \text{N}_2 + 4\text{H}_2\text{O}$ \text{If 11.1 \text{ mol } \text{N}_2\text{H}_4 is mixed with 1.4 \text{ mol } \text{H}_2\text{O}_2, calculate the following:} $\text{Part 3 of 4: How much } \text{N}_2 \text{ is formed? Answer in units of mol.}$ $\text{Part 4 of 4: How much } \text{H}_2\text{O} \text{ is formed? Answer in units of mol.}$

Studdy Solution

STEP 1

What is this asking? Given a chemical reaction, if we mix a certain amount of $\text{N}_2\text{H}_4$ and $\text{H}_2\text{O}_2$ , how much $\text{N}_2$ and $\text{H}_2\text{O}$ will we get? Watch out! We need to figure out which reactant is limiting the reaction before we can calculate how much product is formed.

STEP 2

1. Find the Limiting Reactant
2. Calculate $\text{N}_2$ Formed
3. Calculate $\text{H}_2\text{O}$ Formed

STEP 3

We're given $11.1 \text{ mol}$ of $\text{N}_2\text{H}_4$ and $1.4 \text{ mol}$ of $\text{H}_2\text{O}_2$ .
The balanced equation tells us that we need two times the amount of $\text{H}_2\text{O}_2$ compared to $\text{N}_2\text{H}_4$ .
Let's see how much $\text{H}_2\text{O}_2$ we would need to react with all the $\text{N}_2\text{H}_4$ .

STEP 4

We have $11.1 \text{ mol}$ of $\text{N}_2\text{H}_4$ .
Since we need twice the $\text{H}_2\text{O}_2$ , we would need $2 \cdot 11.1 \text{ mol} = \boldsymbol{22.2 \text{ mol}}$ of $\text{H}_2\text{O}_2$ .

STEP 5

Uh oh!
We only have $1.4 \text{ mol}$ of $\text{H}_2\text{O}_2$ , but we need $\boldsymbol{22.2 \text{ mol}}$ to react with all the $\text{N}_2\text{H}_4$ .
This means $\text{H}_2\text{O}_2$ is our limiting reactant.
It's going to run out first and stop the reaction.

STEP 6

The balanced equation tells us that for every $2 \text{ mol}$ of $\text{H}_2\text{O}_2$ used, we get $1 \text{ mol}$ of $\text{N}_2$ .
This is a mole ratio!

STEP 7

We have $1.4 \text{ mol}$ of $\text{H}_2\text{O}_2$ .
So, we can calculate the moles of $\text{N}_2$ formed: $\frac{1.4 \text{ mol H}_2\text{O}_2}{1} \cdot \frac{1 \text{ mol N}_2}{2 \text{ mol H}_2\text{O}_2} = \boldsymbol{0.7 \text{ mol N}_2}$ .

STEP 8

The balanced equation tells us that for every $2 \text{ mol}$ of $\text{H}_2\text{O}_2$ used, we get $4 \text{ mol}$ of $\text{H}_2\text{O}$ .
Another mole ratio!

STEP 9

We have $1.4 \text{ mol}$ of $\text{H}_2\text{O}_2$ .
So, we can calculate the moles of $\text{H}_2\text{O}$ formed: $\frac{1.4 \text{ mol H}_2\text{O}_2}{1} \cdot \frac{4 \text{ mol H}_2\text{O}}{2 \text{ mol H}_2\text{O}_2} = \boldsymbol{2.8 \text{ mol H}_2\text{O}}$ .

STEP 10

We will produce $\boldsymbol{0.7 \text{ mol}}$ of $\text{N}_2$ and $\boldsymbol{2.8 \text{ mol}}$ of $\text{H}_2\text{O}$ .

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Studdy Solution

STEP 1

STEP 2

1. Find the Limiting Reactant
2. Calculate $\text{N}_2$ Formed
3. Calculate $\text{H}_2\text{O}$ Formed

STEP 3

STEP 4

We have $11.1 \text{ mol}$ of $\text{N}_2\text{H}_4$ .
Since we need twice the $\text{H}_2\text{O}_2$ , we would need $2 \cdot 11.1 \text{ mol} = \boldsymbol{22.2 \text{ mol}}$ of $\text{H}_2\text{O}_2$ .

STEP 5

Uh oh!
We only have $1.4 \text{ mol}$ of $\text{H}_2\text{O}_2$ , but we need $\boldsymbol{22.2 \text{ mol}}$ to react with all the $\text{N}_2\text{H}_4$ .
This means $\text{H}_2\text{O}_2$ is our limiting reactant.
It's going to run out first and stop the reaction.

STEP 6

The balanced equation tells us that for every $2 \text{ mol}$ of $\text{H}_2\text{O}_2$ used, we get $1 \text{ mol}$ of $\text{N}_2$ .
This is a mole ratio!

STEP 7

We have $1.4 \text{ mol}$ of $\text{H}_2\text{O}_2$ .
So, we can calculate the moles of $\text{N}_2$ formed: $\frac{1.4 \text{ mol H}_2\text{O}_2}{1} \cdot \frac{1 \text{ mol N}_2}{2 \text{ mol H}_2\text{O}_2} = \boldsymbol{0.7 \text{ mol N}_2}$ .

STEP 8

The balanced equation tells us that for every $2 \text{ mol}$ of $\text{H}_2\text{O}_2$ used, we get $4 \text{ mol}$ of $\text{H}_2\text{O}$ .
Another mole ratio!

STEP 9

We have $1.4 \text{ mol}$ of $\text{H}_2\text{O}_2$ .
So, we can calculate the moles of $\text{H}_2\text{O}$ formed: $\frac{1.4 \text{ mol H}_2\text{O}_2}{1} \cdot \frac{4 \text{ mol H}_2\text{O}}{2 \text{ mol H}_2\text{O}_2} = \boldsymbol{2.8 \text{ mol H}_2\text{O}}$ .

STEP 10

We will produce $\boldsymbol{0.7 \text{ mol}}$ of $\text{N}_2$ and $\boldsymbol{2.8 \text{ mol}}$ of $\text{H}_2\text{O}$ .

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Studdy Solution

STEP 1

STEP 2

1. Find the Limiting Reactant2. Calculate N2 \text{N}_2 N2​ Formed3. Calculate H2O \text{H}_2\text{O} H2​O Formed

STEP 3

STEP 4

STEP 5

STEP 6

The balanced equation tells us that for every 2 mol 2 \text{ mol} 2 mol of H2O2 \text{H}_2\text{O}_2 H2​O2​ used, we get 1 mol 1 \text{ mol} 1 mol of N2 \text{N}_2 N2​.This is a **mole ratio**!

STEP 7

STEP 8

The balanced equation tells us that for every 2 mol 2 \text{ mol} 2 mol of H2O2 \text{H}_2\text{O}_2 H2​O2​ used, we get 4 mol 4 \text{ mol} 4 mol of H2O \text{H}_2\text{O} H2​O.Another **mole ratio**!

STEP 9

STEP 10

We will produce 0.7 mol \boldsymbol{0.7 \text{ mol}} 0.7 mol of N2 \text{N}_2 N2​ and 2.8 mol \boldsymbol{2.8 \text{ mol}} 2.8 mol of H2O \text{H}_2\text{O} H2​O.

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1. Find the Limiting Reactant
2. Calculate $\text{N}_2$ Formed
3. Calculate $\text{H}_2\text{O}$ Formed

The balanced equation tells us that for every $2 \text{ mol}$ of $\text{H}_2\text{O}_2$ used, we get $1 \text{ mol}$ of $\text{N}_2$ .
This is a mole ratio!

The balanced equation tells us that for every $2 \text{ mol}$ of $\text{H}_2\text{O}_2$ used, we get $4 \text{ mol}$ of $\text{H}_2\text{O}$ .
Another mole ratio!

We will produce $\boldsymbol{0.7 \text{ mol}}$ of $\text{N}_2$ and $\boldsymbol{2.8 \text{ mol}}$ of $\text{H}_2\text{O}$ .