Propane burns in oxygen to produce carbon dioxide and water what is the percent yeild

The percentage yield is 78.2g

Explanation:

Given, mass of propane = 42.8 g , sufficient O2 percent yield = 61.0 % yield.

Reaction - C3H8(g)+5O2(g)------> 3CO2(g)+4H2O(g)

First we need to calculate the moles of propane

Moles of propane = g.mol-1

                            = 0.971 moles

So, moles of CO2 from the moles of propane

1 mole of C3H8(g) = 3 moles of CO2(g)

So, 0.971 moles of C3H8(g) = ?

= 2.913 moles of CO2

So theoretical yield = 2.913 moles 44.0 g/mol

                               = 128.2 g

So, the actual mass of CO2 = percent yield  theoretical yield / 100 %

                                         = 61.0 %  128.2 g / 100 %

                                         = 78.2 g

the mass of CO2 that can be produced if the reaction of 42.8 g of propane and sufficient oxygen has a 61.0 % yield is 78.2 g

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