Determine how many grams of water are produced when burning 1.71 g of hexane, C6H14(l), as a component of gasoline in automobile engines. Assume excess oxygen is present.

2.51 g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below below:

2C6H14 + 19O2 —> 12CO2 + 14H2O

Next, we shall determine the mass of hexane (C6H14) that reacted and the mass of water (H2O) produce from the balanced equation. This is illustrated below:

Molar mass of C6H14 = (12×6) + (14×1)

= 72 + 14

= 86 g/mol

Mass of C6H14 from the balanced equation = 2 × 86 = 172 g

Molar mass of H2O = (2×1) + 16

= 2 + 16

= 18 g/mol

Mass of H2O from the balanced equation = 14 × 18 = 252 g

From the balanced equation above,

172 g of C6H14 reacted to produce 252 g of H2O.

Finally, we shall determine the mass of water produced by burning 1.71 g of hexane, C6H14. This is illustrated below:

From the balanced equation above,

172 g of C6H14 reacted to produce 252 g of H2O.

Therefore, 1.71 g of C6H14 will react to produce = (1.71 × 252) / 172 = 2.51 g of H2O.

Therefore, 2.51 g of H2O were obtained from the reaction.

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