Question

4. Bob was cooking and used up 1.00 kg of CH. during the process. How many klog were produced? Show your work, round your answer to three significant figu units (10 points)? three significant figures, and include C3H8 +502 → 3C02 + 4H20 If Bob was doing the cooking outside where there was plenty of Oz gas available, what was the limiting reactant? If during this cooking, he produced 2.45 kg of CO2, what was his percent yield?

Answer 1

Answer:

1. Here molar mass of C₃H₈ = 44.1 g/mol

Moles of C₃H₈ = 1000/44.1 = 22.67 moles

So, one mole of C₃H₈ will produce three moles of CO₂.

Hence, the number of moles of CO₂ forms = 68.02 moles

Now the molar mass of CO₂ = 44.01 g/mol

Amount of CO₂ formed = 2.994 Kg

2. As there are only two reactants available in this combustion process. The limiting reactant is C₃H₈.

3. The % yield of CO₂ formed = (Actual/Theoretical) x 100 = (2.45/2.994) x 100 = 81.83 %

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