The combustion of hexyne, C6H10 is represented by the given equation:C6H10 + O2 --> CO2 + H2OIf 403 grams of C6H10 is available. If all C6H10 will be fully consumed, how many grams of CO2 will be produced for complete combustion assuming no excess O2 is supplied?Write your final answer in two decimal places.Use the following mass numbers:C - 12 g/molH - 1 g/molO - 16 g/mol

To solve this problem, we need to follow these steps:

1. Calculate the molar mass of C6H10, CO2, and O2.
2. Use stoichiometry to find the ratio of moles of C6H10 to CO2.
3. Use the ratio to calculate the mass of CO2 produced.

Step 1: Calculate the molar mass of C6H10, CO2, and O2.

For C6H10: C6H10 = (612 g/mol) + (101 g/mol) = 72 g/mol + 10 g/mol = 82 g/mol

For CO2: CO2 = (112 g/mol) + (216 g/mol) = 12 g/mol + 32 g/mol = 44 g/mol

Step 2: Use stoichiometry to find the ratio of moles of C6H10 to CO2.

From the balanced chemical equation, we can see that 1 mole of C6H10 produces 6 moles of CO2.

Step 3: Use the ratio to calculate the mass of CO2 produced.

First, we need to find out how many moles of C6H10 we have: 403 g / 82 g/mol = 4.92 moles of C6H10

Since 1 mole of C6H10 produces 6 moles of CO2, 4.92 moles of C6H10 will produce: 4.92 moles * 6 = 29.52 moles of CO2

Finally, we convert moles of CO2 to grams: 29.52 moles * 44 g/mol = 1298.88 g

So, 403 grams of C6H10 will produce approximately 1298.88 grams of CO2.

To solve this problem, we need to follow these steps:

1. Calculate the molar mass of C6H10, CO2, and O2.
2. Use stoichiometry to find the ratio of moles of C6H10 to CO2.
3. Use the ratio to calculate the mass of CO2 produced.

Step 1: Calculate the molar mass of C6H10, CO2, and O2.

• The molar mass of C6H10 is (612 g/mol for C) + (101 g/mol for H) = 72 g/mol + 10 g/mol = 82 g/mol.
• The molar mass of CO2 is (112 g/mol for C) + (216 g/mol for O) = 12 g/mol + 32 g/mol = 44 g/mol.

Step 2: Use stoichiometry to find the ratio of moles of C6H10 to CO2.

• From the balanced chemical equation, we can see that 1 mole of C6H10 produces 6 moles of CO2.

Step 3: Use the ratio to calculate the mass of CO2 produced.

• First, we need to find out how many moles of C6H10 are in 403 grams. We do this by dividing the mass by the molar mass: 403 g / 82 g/mol = 4.92 moles of C6H10.
• Since 1 mole of C6H10 produces 6 moles of CO2, 4.92 moles of C6H10 will produce 4.92 * 6 = 29.52 moles of CO2.
• Finally, we convert this to grams by multiplying by the molar mass of CO2: 29.52 moles * 44 g/mol = 1298.88 g of CO2.

So, if all C6H10 is fully consumed, 1298.88 grams of CO2 will be produced. Rounded to two decimal places, the answer is 1298.88 grams.