How many grams of carbon dioxide can form when a mixture of 3.60 g ethylene (C2H4) and 2.70 g of oxygen is ignited, assuming complete combustion to form carbon dioxide and water?

I hate to do all of this typing over again; please refer to this limiting reagent problem. Just follow the steps.

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To determine the amount of carbon dioxide (CO2) that can form when ethylene (C2H4) reacts with oxygen (O2), you need to follow these steps:

Step 1: Write the balanced chemical equation for the combustion of ethylene:
C2H4 + O2 -> CO2 + H2O

Step 2: Convert the given mass of ethylene (C2H4) to moles. To do this, divide the given mass of ethylene by its molar mass. The molar mass of ethylene is the sum of the molar masses of carbon (C) and hydrogen (H):
Molar mass of C2H4 = (2 * Molar mass of C) + (4 * Molar mass of H)

Step 3: Convert the given mass of oxygen (O2) to moles. To do this, divide the given mass of oxygen by its molar mass. The molar mass of oxygen is:
Molar mass of O2 = 2 * Molar mass of O

Step 4: Determine the limiting reactant. Compare the calculated number of moles of C2H4 and O2. The reactant that produces fewer moles of CO2 is the limiting reactant. Let's assume ethylene (C2H4) is the limiting reactant.

Step 5: Use the coefficients in the balanced equation to determine the number of moles of CO2 produced. From the balanced equation, you can see that 1 mole of C2H4 produces 1 mole of CO2.

Step 6: Convert the moles of CO2 to grams. Multiply the number of moles of CO2 by its molar mass. The molar mass of CO2 is the sum of the molar masses of carbon (C) and oxygen (O):
Molar mass of CO2 = (Molar mass of C) + (2 * Molar mass of O)

Finally, perform the calculations:

Step 1: C2H4 + O2 -> CO2 + H2O

Step 2: Moles of C2H4 = (3.60 g C2H4) / (Molar mass of C2H4)

Step 3: Moles of O2 = (2.70 g O2) / (Molar mass of O2)

Step 4: Assuming C2H4 is the limiting reactant, proceed with that.

Step 5: Moles of CO2 produced = Moles of C2H4

Step 6: Mass of CO2 = (Moles of CO2 produced) * (Molar mass of CO2)

By following these steps, you can calculate the grams of carbon dioxide formed when the given mixture of ethylene and oxygen is ignited.

To determine the number of grams of carbon dioxide (CO2) formed, we first need to balance the chemical equation for the combustion of ethylene:

C2H4 + O2 → CO2 + H2O

To balance the equation, we have:

C2H4 + 3 O2 → 2 CO2 + 2 H2O

From the balanced equation, we can see that 1 mol of ethylene (C2H4) reacts with 3 mol of oxygen (O2) to produce 2 mol of carbon dioxide (CO2).

Step 1: Convert grams of ethylene (C2H4) to moles
Using the molar mass of ethylene (C2H4), which is approximately 28.05 g/mol, we have:

3.60 g C2H4 x (1 mol C2H4/28.05 g C2H4) = 0.128 mol C2H4

Step 2: Convert grams of oxygen (O2) to moles
Using the molar mass of oxygen (O2), which is approximately 32.00 g/mol, we have:

2.70 g O2 x (1 mol O2/32.00 g O2) = 0.0844 mol O2

Step 3: Determine the limiting reactant
To determine the limiting reactant, we compare the number of moles of each reactant. The reactant with the smaller number of moles is the limiting reactant.

From step 1: Moles of C2H4 = 0.128 mol
From step 2: Moles of O2 = 0.0844 mol

Since we need 3 moles of O2 for every 1 mole of C2H4, we multiply the moles of O2 by 3:

0.0844 mol O2 x (3 mol O2/1 mol C2H4) = 0.253 mol O2

Comparing the moles, we can see that the moles of O2 is smaller than the moles of C2H4. Therefore, oxygen is the limiting reactant.

Step 4: Calculate the moles of CO2 formed
From the balanced chemical equation, we know that 2 moles of CO2 are formed for every 1 mole of ethylene.

Since oxygen is the limiting reactant, all of the oxygen will react completely with the ethylene to form 2 moles of CO2. Therefore, we have:

2 mol CO2 x (0.128 mol C2H4/1 mol C2H4) = 0.256 mol CO2

Step 5: Convert moles of CO2 to grams
Using the molar mass of carbon dioxide (CO2), which is approximately 44.01 g/mol, we have:

0.256 mol CO2 x (44.01 g CO2/1 mol CO2) = 11.3 g CO2

Therefore, approximately 11.3 grams of carbon dioxide (CO2) can form when the mixture of 3.60 g of ethylene (C2H4) and 2.70 g of oxygen (O2) is ignited, assuming complete combustion.