Question

name: _guina takemura_ per: _1_ date: _2/5/26_
\gram to mole\ stoichiometry calculations
show all work, use units on your numbers, and round your answer to the proper number of significant figures.

1. how many moles of co₂ will be produced from 121 grams of o₂?

4fecr₂o₇ + 8k₂co₃ + o₂ → 2fe₂o₃ + 8k₂cro₄ + 8co₂

1. how many moles of sulfur are required to react with 75.0 grams of o₂?

s + o₂ → so₂

1. how many moles of aluminum are required to produce 495.0 grams of h₂?

6naoh + 2al → 2na₃alo₃ + 3h₂

1. how many moles of nitrogen are required to produce 68.5 grams of nh₃?

n₂ + 3h₂ → 2nh₃

1. if the reaction below produces 100. grams of oxygen, how many moles of kcl will be produced?

2kclo₃ → 2kcl + 3o₂

Explanation:

Problem 1:

Step 1: Molar mass of \( O_2 \)

Molar mass of \( O = 16.00 \, g/mol \), so \( O_2 \) molar mass is \( 2 \times 16.00 = 32.00 \, g/mol \).

Step 2: Moles of \( O_2 \)

Moles \( = \frac{mass}{molar \, mass} = \frac{121 \, g}{32.00 \, g/mol} \approx 3.78125 \, mol \).

Step 3: Mole ratio from reaction

From \( 4FeCr_2O_7 + 8K_2CO_3 + O_2
ightarrow 2Fe_2O_3 + 8K_2CrO_4 + 8CO_2 \), ratio \( O_2:CO_2 = 1:8 \).

Step 4: Moles of \( CO_2 \)

Moles \( CO_2 = 3.78125 \, mol \times 8 = 30.25 \, mol \) (rounded to 3 significant figures: 30.3 mol? Wait, 121 has 3 sig figs, 32.00 has 4. So 121/32.00 = 3.78125 (5 sig figs), times 8: 30.25, which with 3 sig figs is 30.3? Wait, no: 121 is 3 sig figs, so 121/32.00 = 3.78 (3 sig figs? Wait, 32.00 is 4, so 121/32.00 = 3.78125, then times 8: 30.25, which should be 30.3? Wait, no, 121 is 3 sig figs, so the answer should have 3. Wait, 121 g (3 sig figs), molar mass 32.00 (4), so moles of \( O_2 \) is 121/32.00 = 3.78125 (we can keep more digits for intermediate steps). Then mole ratio 1:8, so 3.78125 8 = 30.25, which rounds to 30.3 mol? Wait, no, 121 is 3 sig figs, so 30.3? Wait, 3.78125 8 = 30.25, which is 30.3 when rounded to 3 sig figs. Wait, but let's check again.

Wait, 121 g \( O_2 \):

Moles \( O_2 = 121 g / 32.00 g/mol = 3.78125 mol \)

From reaction, 1 mol \( O_2 \) produces 8 mol \( CO_2 \), so moles \( CO_2 = 3.78125 mol * 8 = 30.25 mol \approx 30.3 mol \) (3 sig figs).

Step 1: Molar mass of \( O_2 \)

\( O_2 \) molar mass \( = 32.00 \, g/mol \).

Step 2: Moles of \( O_2 \)

Moles \( = \frac{75.0 \, g}{32.00 \, g/mol} = 2.34375 \, mol \).

Step 3: Mole ratio (from \( S + O_2

ightarrow SO_2 \), ratio \( S:O_2 = 1:1 \))
Moles of \( S \) = moles of \( O_2 = 2.34375 \, mol \approx 2.34 \, mol \) (3 sig figs, since 75.0 has 3).

Step 1: Molar mass of \( H_2 \)

Molar mass \( H_2 = 2.016 \, g/mol \).

Step 2: Moles of \( H_2 \)

Moles \( = \frac{495.0 \, g}{2.016 \, g/mol} \approx 245.5357 \, mol \).

Step 3: Mole ratio (from \( 6NaOH + 2Al

ightarrow 2Na_3AlO_3 + 3H_2 \), ratio \( Al:H_2 = 2:3 \))
Moles of \( Al = \frac{2}{3} \times 245.5357 \, mol \approx 163.69 \, mol \approx 164 \, mol \) (3 sig figs? Wait, 495.0 has 4, 2.016 has 4, so let's calculate more precisely.

\( 495.0 / 2.016 = 245.535714... \)

Then \( (2/3) 245.535714 = 163.690476... \approx 164 \, mol \) (if 3 sig figs, but 495.0 has 4, so maybe 163.7 mol? Wait, 495.0 is 4 sig figs, 2.016 is 4, so the answer should have 4? Wait, 245.5357 (2/3) = 163.6905, which is 163.7 mol (4 sig figs).

Answer:

30.3 mol (or 30.25 if 4 sig figs, but 121 has 3, so 30.3)

Problem 2: