Question

The reaction of 6.30 g of carbon with excess O₂ yields 11.9 g of CO₂. What is the percent yield of this reaction?

Answer 1

answer)

first we have to understand what is percentage yield,

percentage yield; this is referrs to the ratio of actual yield to theoretical yield.

percentage yield = (actual yield / theoretical yield ) * 100%

we can find the percentage yield in the following way,

first write the balanced equation for the given reaction. here carbon reacts with oxygen molecule to form carbon dioxide, the balanced equation is,

C + O2 CO2

then we have to converts the given datas in the question to moles. this can be done by the equation

moles = grams / atomic mass

here mass of carbon in grams = 6.30g (given)

atomic mass of carbon = 12 g

so, the moles of carbon = 6.30 g / 12 g

= 0.525 moles

then , using the coefficients in the balanced chemical equation converts moles of carbon to moles of carbon dioxide(product).

this can be done in following way,

0.525 mols carbon * (1 mol of carbon dioxide / 1mol of carbon) = 0.525 * (1/1)

= 0.525 moles of CO2

the next step is converting moles of CO2 in to grams. this can be done by using the following formula,

grams = given mols * molar mass

here, moles of CO2 = 0.525 mols

molar mass of CO2 = 44 g

so, grams = 0.525 mols * 44 g

= 23.1 g CO2

this is the theoretical yield of CO2

theoretical yield of CO2 = 23.1 g

actual yield of CO2 is given in the problem,

actual yield of CO2 = 11.9 g

so we can calculate the percentage yield by the following equation,

percentage yield = (actual yield / theoretical yield) * 100

= (11.9 g / 23.1 g) *100

= 0.515 * 100

= 51.5%

percentage yield of CO2 = 51.5%