Question

What is the mass of water formed when 1.6 g of methane gas is completely burnt in excess oxygen?

• A. 1.8 g
• B. 2.4 g
• C. 3.2 g
• D. 3.6 g
• E. 4.8 g

Hint:

Always use the mole ratio from the balanced equation to relate the moles of reactants and products. Then use the molar masses to convert between grams and moles.

Answer 1

The Correct Option is D

The combustion of methane (CH₄) in excess oxygen (O₂) forms carbon dioxide (CO₂) and water (H₂O). The balanced chemical equation for the combustion reaction is: \[ {CH}_4 (g) + 2 {O}_2 (g) \rightarrow {CO}_2 (g) + 2 {H}_2{O} (g) \]
From the equation, we can see that 1 mole of methane produces 2 moles of water.
Step 1: Calculate the molar mass of methane (CH₄) and water (H₂O):
- Molar mass of CH₄ = \( 12 + (4 \times 1) = 16 \, {g/mol} \)
- Molar mass of H₂O = \( (2 \times 1) + 16 = 18 \, {g/mol} \)
Step 2: Convert the mass of methane to moles:
We are given 1.6 g of methane. To calculate the number of moles of methane: \[ {Moles of CH₄} = \frac{1.6 \, {g}}{16 \, {g/mol}} = 0.1 \, {mol} \]
Step 3: Use the mole ratio from the balanced equation to find the moles of water:
From the balanced equation, 1 mole of methane produces 2 moles of water. Therefore, 0.1 mole of methane will produce: \[ {Moles of H₂O} = 0.1 \, {mol} \times 2 = 0.2 \, {mol} \] Step 4: Calculate the mass of water:
Now, to find the mass of water, multiply the moles of water by its molar mass: \[ {Mass of H₂O} = 0.2 \, {mol} \times 18 \, {g/mol} = 3.6 \, {g} \] Thus, the correct answer is option (D), 3.6 g.