Ethylene [tex]\( \left( C_2H_4 \right) \)[/tex] is the starting material for the preparation of polyethylene. Although typically made during the processing of petroleum, ethylene occurs naturally as a fruit-ripening hormone and as a component of natural gas.

The heat of reaction for the combustion of [tex]\( C_2H_4 \)[/tex] is [tex]\( -1411 \, \frac{\text{kJ}}{\text{mol}} \)[/tex].

How many grams of [tex]\( C_2H_4 \)[/tex] must burn to give 98.3 kJ of heat? Round your answer to 3 significant figures.

[tex]\( \square \) \, \text{g} \, C_2H_4[/tex]

To find how many grams of ethylene ([tex]\(C_2H_4\)[/tex]) must burn to give 98.3 kJ of heat, follow these steps:

1. Understand the heat of reaction: The heat of reaction for the combustion of [tex]\(C_2H_4\)[/tex] is [tex]\(-1411 \, \text{kJ/mol}\)[/tex]. This means that when one mole of [tex]\(C_2H_4\)[/tex] burns, it releases 1411 kJ of heat.

2. Determine the moles of [tex]\(C_2H_4\)[/tex] needed: Since we want to produce 98.3 kJ of heat, we need to calculate how many moles of [tex]\(C_2H_4\)[/tex] are required for this energy release:

[tex]\[
\text{Moles of } C_2H_4 = \frac{\text{Desired Heat}}{\text{Heat of Reaction}} = \frac{98.3 \, \text{kJ}}{1411 \, \text{kJ/mol}}
\][/tex]

This gives approximately 0.06967 moles of [tex]\(C_2H_4\)[/tex].

3. Calculate the grams of [tex]\(C_2H_4\)[/tex] required: Next, we'll use the molar mass of [tex]\(C_2H_4\)[/tex] to convert moles to grams. The molar mass of [tex]\(C_2H_4\)[/tex] is 28.05 g/mol. Therefore, the mass in grams of [tex]\(C_2H_4\)[/tex] needed is:

[tex]\[
\text{Grams of } C_2H_4 = \text{Moles of } C_2H_4 \times \text{Molar Mass of } C_2H_4 = 0.06967 \, \text{mol} \times 28.05 \, \text{g/mol}
\][/tex]

This results in approximately 1.954 grams of [tex]\(C_2H_4\)[/tex].

4. Rounding the answer: Finally, round the result to three significant figures to match the precision of the given data.

So, approximately 1.954 grams of [tex]\(C_2H_4\)[/tex] must burn to produce 98.3 kJ of heat.