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How many grams of H2O will be produced by the combustion of 16 grams of C2H4?

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Answer:

The balanced chemical equation for the combustion of C2H4 (ethylene) is:

C2H4 + 3 O2 → 2 CO2 + 2 H2O

From the balanced chemical equation, we can see that 1 mole of C2H4 reacts to produce 2 moles of H2O.

First, we need to convert the given mass of C2H4 to moles using its molar mass. The molar mass of C2H4 is:

2(12.01 g/mol) + 4(1.01 g/mol) = 28.05 g/mol

Now, we can calculate the moles of C2H4:

moles of C2H4 = mass of C2H4 / molar mass of C2H4

moles of C2H4 = 16 g / 28.05 g/mol

moles of C2H4 ≈ 0.570 mol (rounded to three decimal places)

According to the balanced chemical equation, 1 mole of C2H4 produces 2 moles of H2O.

moles of H2O = 2 * moles of C2H4

moles of H2O = 2 * 0.570 mol

moles of H2O ≈ 1.140 mol (rounded to three decimal places)

Finally, we can convert the moles of H2O to grams using the molar mass of water (H2O), which is 18.015 g/mol.

mass of H2O = moles of H2O * molar mass of H2O

mass of H2O = 1.140 mol * 18.015 g/mol

mass of H2O ≈ 20.53 g (rounded to two decimal places)

So, approximately 20.53 grams of H2O will be produced by the combustion of 16 grams of C2H4.

User Stuart Mitchell
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