Could someone please walk me through step by step on how to do this.

Propane (C3H8) burns in oxygen to produce carbon dioxide and water.
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)

Calculate the mass of CO2 that can be produced if the reaction of 48.2 g propane and sufficient oxygen has a 62.3% yield.

In Grams

You have the equation. Follow the steps in a sample problem I posted here.

http://www.jiskha.com/science/chemistry/stoichiometry.html

When you have the final mass CO2, multiply by 0.623 to correct for it not being a 100% yield.

The half-life of I-131 is 8 days. The lab has a sample of I-131 weighing 150g. How much would be left after 40 days?

-12ml, 60cm, and 4.0g. Ehat is the objects density?

To calculate the mass of CO2 produced, you need to follow these steps:

Step 1: Write and balance the chemical equation:
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)

Step 2: Convert the mass of propane to moles:
To do this, you need to find the molar mass of propane (C3H8).

C: 12.01 g/mol x 3 = 36.03 g/mol
H: 1.01 g/mol x 8 = 8.08 g/mol

Add the molar masses:
36.03 g/mol + 8.08 g/mol = 44.11 g/mol

Now divide the given mass of propane (48.2 g) by the molar mass to get moles:
48.2 g / 44.11 g/mol ≈ 1.094 mol

Step 3: Determine the limiting reactant:
Since the question states that sufficient oxygen is provided, the limiting reactant will be propane.

Step 4: Calculate the theoretical yield of CO2:
From the balanced equation, you can see that 1 mole of propane produces 3 moles of CO2.
Therefore, the moles of CO2 produced will be 1.094 moles x 3 = 3.282 moles.

Step 5: Convert moles of CO2 to grams:
To do this, you need to find the molar mass of CO2 (carbon dioxide).
C: 12.01 g/mol
O: 16.00 g/mol x 2 = 32.00 g/mol

Add the molar masses:
12.01 g/mol + 32.00 g/mol = 44.01 g/mol

Now multiply the moles of CO2 by the molar mass to get the mass:
3.282 moles x 44.01 g/mol ≈ 144.38 g

Step 6: Calculate the actual yield:
The actual yield is given as 62.3% of the theoretical yield.
Multiply the theoretical yield (144.38 g) by 0.623 to get the actual yield:
144.38 g x 0.623 ≈ 89.97 g (rounded to 2 decimal places)

Therefore, the mass of CO2 produced in this reaction, with a 62.3% yield, is approximately 89.97 grams.