Question

5 multiple choice 1 point
potassium oxide (k₂o) is often used in agriculture as a fertilizer. which represents a balanced chemical equation for producing potassium oxide?
4k + o₂ → 2k₂o
k + 2o₂ → k2o
k + o₂ → k₂o

6 multiple choice 1 point
hydrogen and oxygen (o₂) react to form liquid water. if there are 10 grams of oxygen reacting with hydrogen, how many moles of oxygen would be in the resulting water product?
0.55 moles
0.31 moles
1.12 moles
2.34 moles

7 multiple choice 1 point
combustion reactions always produce...
carbon monoxide and water.
carbon dioxide and water.
pure oxygen and hydrogen.
pure oxygen and water.

Explanation:

Question 5

Step1: Recall Balancing Rules

A balanced chemical equation has equal numbers of each atom on both sides. For \( \text{K}_2\text{O} \) production from \( \text{K} \) and \( \text{O}_2 \):

• **Option 1 (\( 4\text{K} + \text{O}_2

ightarrow 2\text{K}_2\text{O} \))**:

• K: Left = 4, Right = \( 2\times2 = 4 \).
• O: Left = 2, Right = \( 2\times1 = 2 \). Balanced.
• **Option 2 (\( \text{K} + 2\text{O}_2

ightarrow \text{K}_2\text{O} \))**:

• K: Left = 1, Right = 2. Unbalanced.
• O: Left = 4, Right = 1. Unbalanced.
• **Option 3 (\( \text{K} + \text{O}_2

ightarrow \text{K}_2\text{O} \))**:

• K: Left = 1, Right = 2. Unbalanced.
• O: Left = 2, Right = 1. Unbalanced.

Step1: Molar Mass of \( \text{O}_2 \)

Molar mass of \( \text{O}_2 = 2\times16 = 32 \, \text{g/mol} \).

Step2: Moles of \( \text{O}_2 \)

Moles of \( \text{O}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{10 \, \text{g}}{32 \, \text{g/mol}} \approx 0.3125 \, \text{mol} \).

Step3: Reaction Stoichiometry

Reaction: \( 2\text{H}_2 + \text{O}_2
ightarrow 2\text{H}_2\text{O} \). 1 mol \( \text{O}_2 \) produces 2 mol \( \text{H}_2\text{O} \), but moles of O in \( \text{H}_2\text{O} \): 1 mol \( \text{O}_2 \) (2 mol O) → 2 mol \( \text{H}_2\text{O} \) (2 mol O). So moles of O in \( \text{H}_2\text{O} \) = moles of \( \text{O}_2 \) (since 1 \( \text{O}_2 \) has 2 O, and 1 \( \text{H}_2\text{O} \) has 1 O; ratio O in \( \text{O}_2 \) to O in \( \text{H}_2\text{O} \) is 1:1? Wait, no: \( \text{O}_2 \) (2 O atoms) → 2 \( \text{H}_2\text{O} \) (2 O atoms). So moles of O in \( \text{H}_2\text{O} \) = moles of \( \text{O}_2 \times 2 \)? Wait, no: \( \text{O}_2 \) is diatomic. Let's re-express:
Moles of \( \text{O}_2 = 10/32 \approx 0.3125 \, \text{mol} \). Each \( \text{O}_2 \) has 2 O atoms, so moles of O atoms = \( 0.3125 \times 2 = 0.625 \)? No, wait the question is "moles of oxygen in water". Wait, water is \( \text{H}_2\text{O} \), so each \( \text{H}_2\text{O} \) has 1 O atom. The reaction: \( \text{O}_2 \) (2 O) → 2 \( \text{H}_2\text{O} \) (2 O). So moles of O in \( \text{H}_2\text{O} \) = moles of \( \text{O}_2 \times 2 \)? No, wait moles of \( \text{O}_2 \) is 0.3125 mol. Each \( \text{O}_2 \) molecule has 2 O atoms, so total O atoms moles = 0.3125 × 2 = 0.625? But the options have 0.31, 0.55, etc. Wait, maybe I misread: the question is "moles of oxygen in the resulting water". Wait, water is \( \text{H}_2\text{O} \), so moles of O in water = moles of water. From reaction, 1 mol \( \text{O}_2 \) produces 2 mol \( \text{H}_2\text{O} \). So moles of \( \text{H}_2\text{O} \) = 2 × moles of \( \text{O}_2 \) = 2 × (10/32) ≈ 0.625? No, that's not matching. Wait, maybe the question is moles of O atoms in water, not moles of \( \text{O}_2 \) in water. Wait, the options: 0.31 is close to 10/32 ≈ 0.3125. Wait, maybe the question is moles of \( \text{O}_2 \) reacted, but no—wait, the question says "moles of oxygen would be in the resulting water product". Wait, water has O atoms, not \( \text{O}_2 \). But maybe the question has a typo, or I misinterpret. Wait, let's recalculate:

Molar mass of \( \text{O}_2 = 32 \, \text{g/mol} \). Moles of \( \text{O}_2 = 10 / 32 ≈ 0.3125 \, \text{mol} \). In the reaction \( 2\text{H}_2 + \text{O}_2
ightarrow 2\text{H}_2\text{O} \), 1 mol \( \text{O}_2 \) produces 2 mol \( \text{H}_2\text{O} \). Each \( \text{H}_2\text{O} \) has 1 O atom, so moles of O in \( \text{H}_2\text{O} \) = moles of \( \text{H}_2\text{O} \) = 2 × moles of \( \text{O}_2 \) = 2 × 0.3125 ≈ 0.625? No, that's not an option. Wait, maybe the question is moles of \( \text{O}_2 \) reacted, but the options have 0.31, which is ~10/32. Wait, maybe the question is "moles of oxygen atoms" but the options are labeled as moles of \( \text{O}_2 \)? No, that doesn't make sense. Wait, maybe I made a mistake. Wait, the options: 0.31 is close to 10/32 ≈ 0.3125. So maybe the question is moles of \( \text{O}_2 \) reacted, which is 0.31 mol (rounded).

Combustion reactions (of hydrocarbons or organic compounds with \( \text{O}_2 \)) produce \( \text{CO}_2 \) and \( \text{H}_2\text{O} \) (complete combustion). Incomplete combustion may produce CO, but the standard result is \( \text{CO}_2 \) and \( \text{H}_2\text{O} \).

Answer:

A. \( 4\text{K} + \text{O}_2
ightarrow 2\text{K}_2\text{O} \)

Question 6