Stoichiometry practice problems from a Quizizz worksheet.

Here are the step-by-step solutions for the stoichiometry problems shown in your worksheet.

Question 1

Problem: $2\text{NaClO}_3 \rightarrow 2\text{NaCl} + 3\text{O}_2$. How many grams of $\text{O}_2$ are produced from 12.00 moles of $\text{NaClO}_3$?

1. Find the mole ratio: Look at the coefficients in the balanced equation. For every 2 moles of $\text{NaClO}_3$, you get 3 moles of $\text{O}_2$.
2. Calculate moles of $\text{O}_2$:
$$12.00 \text{ moles } \text{NaClO}_3 \times \frac{3 \text{ moles } \text{O}_2}{2 \text{ moles } \text{NaClO}_3} = 18.00 \text{ moles } \text{O}_2$$
3. Convert to grams: The molar mass of $\text{O}_2$ is $16.00 \times 2 = 32.00 \text{ g/mol}$.
$$18.00 \text{ moles} \times 32.00 \text{ g/mol} = 576 \text{ grams}$$

Correct Option: A (576 g of O₂)

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Question 2

Problem: $\text{Cl}_2 + 2\text{KBr} \rightarrow \text{Br}_2 + 2\text{KCl}$. How many grams of $\text{KCl}$ are produced from 356 g of $\text{KBr}$?

1. Find Molar Masses:
* $\text{KBr}$ (Potassium Bromide): $39.1 + 79.9 = 119.0 \text{ g/mol}$
* $\text{KCl}$ (Potassium Chloride): $39.1 + 35.5 = 74.6 \text{ g/mol}$
2. Convert grams of KBr to moles:
$$356 \text{ g} / 119.0 \text{ g/mol} \approx 2.99 \text{ moles KBr}$$
3. Use the mole ratio: The equation shows a 2:2 ratio between $\text{KBr}$ and $\text{KCl}$, which simplifies to 1:1. So, you produce 2.99 moles of $\text{KCl}$.
4. Convert moles of KCl to grams:
$$2.99 \text{ moles} \times 74.6 \text{ g/mol} \approx 223 \text{ grams}$$

Correct Option: A (223 g)

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Question 3

Problem: $\text{Fe}_2\text{O}_3 + 3\text{H}_2 \rightarrow 2\text{Fe} + 3\text{H}_2\text{O}$. About how many grams of $\text{H}_2\text{O}$ are produced from 150 grams of $\text{Fe}_2\text{O}_3$?

1. Find Molar Masses:
* $\text{Fe}_2\text{O}_3$: $(55.8 \times 2) + (16.0 \times 3) = 111.6 + 48.0 = 159.6 \text{ g/mol}$
* $\text{H}_2\text{O}$: $(1.0 \times 2) + 16.0 = 18.0 \text{ g/mol}$
2. Convert grams of $\text{Fe}_2\text{O}_3$ to moles:
$$150 \text{ g} / 159.6 \text{ g/mol} \approx 0.94 \text{ moles}$$
3. Use the mole ratio: The ratio of $\text{Fe}_2\text{O}_3$ to $\text{H}_2\text{O}$ is 1:3.
$$0.94 \text{ moles} \times 3 = 2.82 \text{ moles H}_2\text{O}$$
4. Convert to grams:
$$2.82 \text{ moles} \times 18.0 \text{ g/mol} \approx 50.76 \text{ grams}$$
This is closest to 50 grams.

Correct Option: D (50 grams H₂O)

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Question 4

Problem: Same reaction as #1. 12.00 moles of $\text{NaClO}_3$ will produce how many grams of $\text{O}_2$?

*Note: This is the exact same calculation as Question 1.*
1. Moles $\text{O}_2 = 12 \times (3/2) = 18$ moles.
2. Grams $\text{O}_2 = 18 \times 32 = 576$ grams.

Correct Option: A (576 g of O₂)

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Question 5

Problem: $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$. How many moles of hydrogen are needed to react with 2 moles of nitrogen?

1. Look at the coefficients: The balanced equation says 1 mole of Nitrogen ($\text{N}_2$) reacts with 3 moles of Hydrogen ($\text{H}_2$).
2. Calculate: If you have 2 moles of Nitrogen, you need twice as much Hydrogen.
$$2 \text{ moles N}_2 \times 3 = 6 \text{ moles H}_2$$

Correct Option: A (6)

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Question 6

Problem: How do I move from grams to moles?

To convert from mass (grams) to amount (moles), you must divide by the substance's molar mass (found on the periodic table).
Formula: $\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}$

*(Note: Since there are no multiple-choice options visible for this question in the image, the answer is the method described above.)*

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Final Answer:
1. A
2. A
3. D
4. A
5. A
6. Divide by the molar mass