QuestionCalculate the moles of oxygen from the reaction of 0.800 mol of ammonium perchlorate $\left(\mathrm{NH}_{4} \mathrm{ClO}_{4}\right)$ .

Studdy Solution

STEP 1

Assumptions1. The chemical reaction is as follows $\mathrm{NH}_{4} \mathrm{Cl}_{4} \rightarrow \mathrm{}_{} + \mathrm{Cl}_{} + \mathrm{}_{} + \mathrm{H}_{}\mathrm{}$ . The reaction is balanced, meaning the number of atoms of each element on the reactant side equals the number of atoms of each element on the product side.
3. The reaction goes to completion, meaning all of the ammonium perchlorate reacts.
4. We are given that0.800 mol of ammonium perchlorate reacts.

STEP 2

First, we need to balance the chemical equation. The balanced equation is $2\mathrm{NH}_{4} \mathrm{Cl}_{4} \rightarrow \mathrm{}_{2} +2\mathrm{Cl}_{2} +4\mathrm{}_{2} +4\mathrm{H}_{2}\mathrm{}$

STEP 3

From the balanced equation, we can see that2 moles of ammonium perchlorate produce moles of oxygen.

STEP 4

We can set up a ratio to find the moles of oxygen produced by0.800 mol of ammonium perchlorate.
$\frac{4 \, \text{moles of oxygen}}{2 \, \text{moles of ammonium perchlorate}} = \frac{x \, \text{moles of oxygen}}{0.800 \, \text{moles of ammonium perchlorate}}$

STEP 5

Now, we can solve for x, which represents the moles of oxygen.
$x = \frac{4 \, \text{moles of oxygen} \times0.800 \, \text{moles of ammonium perchlorate}}{2 \, \text{moles of ammonium perchlorate}}$

STEP 6

Calculate the moles of oxygen.
$x =1.6 \, \text{moles of oxygen}$ The reaction of0.800 mol of ammonium perchlorate produces1.6 mol of oxygen.

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QuestionCalculate the moles of oxygen from the reaction of 0.800 mol of ammonium perchlorate $\left(\mathrm{NH}_{4} \mathrm{ClO}_{4}\right)$ .

Studdy Solution

STEP 1

STEP 2

First, we need to balance the chemical equation. The balanced equation is $2\mathrm{NH}_{4} \mathrm{Cl}_{4} \rightarrow \mathrm{}_{2} +2\mathrm{Cl}_{2} +4\mathrm{}_{2} +4\mathrm{H}_{2}\mathrm{}$

STEP 3

From the balanced equation, we can see that2 moles of ammonium perchlorate produce moles of oxygen.

STEP 4

We can set up a ratio to find the moles of oxygen produced by0.800 mol of ammonium perchlorate.
$\frac{4 \, \text{moles of oxygen}}{2 \, \text{moles of ammonium perchlorate}} = \frac{x \, \text{moles of oxygen}}{0.800 \, \text{moles of ammonium perchlorate}}$

STEP 5

Now, we can solve for x, which represents the moles of oxygen.
$x = \frac{4 \, \text{moles of oxygen} \times0.800 \, \text{moles of ammonium perchlorate}}{2 \, \text{moles of ammonium perchlorate}}$

STEP 6

Calculate the moles of oxygen.
$x =1.6 \, \text{moles of oxygen}$ The reaction of0.800 mol of ammonium perchlorate produces1.6 mol of oxygen.

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QuestionCalculate the moles of oxygen from the reaction of 0.800 mol of ammonium perchlorate (NH4ClO4)\left(\mathrm{NH}_{4} \mathrm{ClO}_{4}\right)(NH4​ClO4​).

Studdy Solution

STEP 1

STEP 2

First, we need to balance the chemical equation. The balanced equation is2NH4Cl4→2+2Cl2+42+4H22\mathrm{NH}_{4} \mathrm{Cl}_{4} \rightarrow \mathrm{}_{2} +2\mathrm{Cl}_{2} +4\mathrm{}_{2} +4\mathrm{H}_{2}\mathrm{}2NH4​Cl4​→2​+2Cl2​+42​+4H2​

STEP 3

From the balanced equation, we can see that2 moles of ammonium perchlorate produce moles of oxygen.

STEP 4

STEP 5

STEP 6

Calculate the moles of oxygen.x=1.6 moles of oxygenx =1.6 \, \text{moles of oxygen}x=1.6moles of oxygenThe reaction of0.800 mol of ammonium perchlorate produces1.6 mol of oxygen.

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QuestionCalculate the moles of oxygen from the reaction of 0.800 mol of ammonium perchlorate $\left(\mathrm{NH}_{4} \mathrm{ClO}_{4}\right)$ .

First, we need to balance the chemical equation. The balanced equation is $2\mathrm{NH}_{4} \mathrm{Cl}_{4} \rightarrow \mathrm{}_{2} +2\mathrm{Cl}_{2} +4\mathrm{}_{2} +4\mathrm{H}_{2}\mathrm{}$

Calculate the moles of oxygen.
$x =1.6 \, \text{moles of oxygen}$ The reaction of0.800 mol of ammonium perchlorate produces1.6 mol of oxygen.