2C4H10 + 13 02 -- 8 CO2 + 10H20
If i want to produce 50g of H20 using the above combustion reaction, how many mol of C4H10 should i use ?
0.55 mol C4H10
10 mol C4H10
2.77 mol C4H10
3.79 mol C4H10

Option 1. 0.55 mol of C₄H₁₀

Explanation:

The reaction is:

2C₄H₁₀ + 13O₂ → 8 CO₂ + 10H₂O

This is a combustion reaction where carbon dioxide and water are produced.

We convert mass of produced water to moles → 50 g / 18 g/mol = 2.78 moles of water.

The stoichometry states that:

10 moles of water are made by 2 moles of C₄H₁₀

Therefore 2.78 moles of water will be made by (2.78 . 2) / 10 = 0.55 mol of C₄H₁₀

We should use 0.55 moles C4H10 to produce 50 grams of H2O (option 1 is correct)

Explanation:

Step 1: data given

Mass of H2O = 50 grams

Molar mass H2O = 18.02 g/mol

Molar mass C4H10 = 58.12 g/mol

Step 2: The balanced equation

2C4H10 + 13 02 → 8 CO2 + 10H20

Step 3: Calculate moles H2O

Moles H2O = 50.0 grams / 18.02 g/mol

Moles H2O = 2.77 moles

Step 4: Calculate moles C4H10

For 2 moles C4H10 we need 13 moles O2 to produce 8 moles CO2 and 10 moles H2O

For 2.77 moles H2O we need 2.77/5 = 0.55 moles C4H10

We should use 0.55 moles C4H10 to produce 50 grams of H2O (option 1 is correct)

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