Math / Algebra

QuestionFor example, if a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen in following reaction, use stoichiometry to determine the limiting reagent: $4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \longrightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}$

Studdy Solution

STEP 1

What is this asking? We've got a chemical reaction where ammonia and oxygen mix to make new stuff, and we need to figure out which one runs out first! Watch out! Don't forget to convert grams to moles – we're dealing with molecules, not just weight!

STEP 2

1. Calculate moles of reactants
2. Determine the limiting reactant

STEP 3

Alright, let's start by calculating the number of moles of each reactant.
Remember, moles are just a way to count tons of tiny molecules all at once!
We'll use the molar mass to convert from grams to moles.

STEP 4

For ammonia ( $\mathrm{NH}_3$ ), the molar mass is approximately $14.01 + 3 \cdot 1.01 = 17.04 \, \mathrm{g/mol}$ .
We have $\mathrm{2.00 \, g}$ of ammonia, so the number of moles is: $\frac{2.00 \, \mathrm{g}}{17.04 \, \mathrm{g/mol}} \approx 0.117 \, \mathrm{mol}$ So we have about $\mathbf{0.117}$ moles of ammonia ready to react!

STEP 5

Now, let's do the same for oxygen ( $\mathrm{O}_2$ ).
Its molar mass is $2 \cdot 16.00 = 32.00 \, \mathrm{g/mol}$ .
We have $\mathrm{4.00 \, g}$ of oxygen, so the number of moles is: $\frac{4.00 \, \mathrm{g}}{32.00 \, \mathrm{g/mol}} = 0.125 \, \mathrm{mol}$ We've got $\mathbf{0.125}$ moles of oxygen ready to go!

STEP 6

Now for the main event: finding the limiting reactant!
This is the reactant that gets used up first, stopping the reaction in its tracks.
We'll use the mole ratio from the balanced chemical equation to figure this out.

STEP 7

The balanced equation tells us that we need $4$ moles of $\mathrm{NH}_3$ for every $5$ moles of $\mathrm{O}_2$ .
Let's see how many moles of $\mathrm{O}_2$ we'd need to react with all of our $\mathrm{NH}_3$ : $0.117 \, \mathrm{mol} \, \mathrm{NH}_3 \cdot \frac{5 \, \mathrm{mol} \, \mathrm{O}_2}{4 \, \mathrm{mol} \, \mathrm{NH}_3} \approx 0.146 \, \mathrm{mol} \, \mathrm{O}_2$

STEP 8

Uh oh!
We only have $\mathbf{0.125}$ moles of $\mathrm{O}_2$ , but we'd need $\mathbf{0.146}$ moles to react with all the ammonia.
That means oxygen is our limiting reactant!
It's going to run out first, leaving some ammonia leftover.

STEP 9

The limiting reactant is oxygen ( $\mathrm{O}_2$ ).

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QuestionFor example, if a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen in following reaction, use stoichiometry to determine the limiting reagent: $4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \longrightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}$

Studdy Solution

STEP 1

What is this asking? We've got a chemical reaction where ammonia and oxygen mix to make new stuff, and we need to figure out which one runs out first! Watch out! Don't forget to convert grams to moles – we're dealing with molecules, not just weight!

STEP 2

1. Calculate moles of reactants
2. Determine the limiting reactant

STEP 3

Alright, let's start by calculating the number of moles of each reactant.
Remember, moles are just a way to count tons of tiny molecules all at once!
We'll use the molar mass to convert from grams to moles.

STEP 4

STEP 5

STEP 6

Now for the main event: finding the limiting reactant!
This is the reactant that gets used up first, stopping the reaction in its tracks.
We'll use the mole ratio from the balanced chemical equation to figure this out.

STEP 7

STEP 8

Uh oh!
We only have $\mathbf{0.125}$ moles of $\mathrm{O}_2$ , but we'd need $\mathbf{0.146}$ moles to react with all the ammonia.
That means oxygen is our limiting reactant!
It's going to run out first, leaving some ammonia leftover.

STEP 9

The limiting reactant is oxygen ( $\mathrm{O}_2$ ).

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QuestionFor example, if a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen in following reaction, use stoichiometry to determine the limiting reagent: 4NH3+5O2⟶4NO+6H2O4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \longrightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}4NH3​+5O2​⟶4NO+6H2​O

Studdy Solution

STEP 1

What is this asking? We've got a chemical reaction where ammonia and oxygen mix to make new stuff, and we need to figure out which one runs out first! Watch out! Don't forget to convert grams to moles – we're dealing with *molecules*, not just weight!

STEP 2

1. Calculate moles of reactants2. Determine the limiting reactant

STEP 3

Alright, let's **start** by calculating the number of moles of each reactant.Remember, moles are just a way to count *tons* of tiny molecules all at once!We'll use the **molar mass** to convert from grams to moles.

STEP 4

STEP 5

STEP 6

Now for the main event: finding the **limiting reactant**!This is the reactant that gets used up *first*, stopping the reaction in its tracks.We'll use the **mole ratio** from the balanced chemical equation to figure this out.

STEP 7

STEP 8

Uh oh!We only have 0.125\mathbf{0.125}0.125 **moles** of O2\mathrm{O}_2O2​, but we'd need 0.146\mathbf{0.146}0.146 **moles** to react with *all* the ammonia.That means oxygen is our **limiting reactant**!It's going to run out first, leaving some ammonia leftover.

STEP 9

The **limiting reactant** is **oxygen** (O2\mathrm{O}_2O2​).

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QuestionFor example, if a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen in following reaction, use stoichiometry to determine the limiting reagent: $4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \longrightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}$

What is this asking? We've got a chemical reaction where ammonia and oxygen mix to make new stuff, and we need to figure out which one runs out first! Watch out! Don't forget to convert grams to moles – we're dealing with molecules, not just weight!

1. Calculate moles of reactants
2. Determine the limiting reactant

Alright, let's start by calculating the number of moles of each reactant.
Remember, moles are just a way to count tons of tiny molecules all at once!
We'll use the molar mass to convert from grams to moles.

Now for the main event: finding the limiting reactant!
This is the reactant that gets used up first, stopping the reaction in its tracks.
We'll use the mole ratio from the balanced chemical equation to figure this out.

Uh oh!
We only have $\mathbf{0.125}$ moles of $\mathrm{O}_2$ , but we'd need $\mathbf{0.146}$ moles to react with all the ammonia.
That means oxygen is our limiting reactant!
It's going to run out first, leaving some ammonia leftover.

The limiting reactant is oxygen ( $\mathrm{O}_2$ ).