Question

Given the following balanced reaction: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

A.) Given 7.65 kg of C3H8 and 3.14 kg of O2,determine the limiting reagent

B.)Determine the number of moles of CO2 produced

C.)Determine the number of kg of CO2 produced

D.)Determine the number of kg of excess reagent left

E.)Given that only 1.88 kg of CO2 are actually produced what is the percent yeild?

Answer 1

(A)

Moles of C3H8 = 7.65 *103 / 44 = 173.86 mol

moles of O2 = 3.14 * 103 / 32 = 98.125 mol

According to the balanced equation,

1 mol C3H8 = 5 mol O2

So, O2 is limiting reagent.

(B) From balaced equation, 5 mol O2 = 3 mol CO2

then, 98.125 mol O2 = 58.875 mol of CO2 produced

(C) Mass of CO2 produced = 58.875 * 44 = 2590.5 g. = 2.5905 kg

(D)

For 98.125 mol of 1*098.125/3 = 32.71 mol of C3H8 is required.

SO, moles of excess reagent = 173.86 - 32.71 = 141.15 mol

Mass of C3H8 remained = 141.5 * 44 = 6210.6 g. = 6.2106 kg.

(E) % yield of CO2 = (1.88/2.5905) * 100 = 72.57 %