Question

You are provided with 6.0g of C and 100g of O2. calculate the amount of CO2 prepared by reacting them. which of them is the limiting reactant.​

Answer 1

Answer:

C+O2 gives CO2

first convert the mass into moles as moles of C=mass divided by molar mass

moles=6 divided by 12=0.5mol

moles of O2=100 divided by 36=2.7mol

now according to the balanced chemical equation

1mol of C=1mol of CO2, so

0.5mol of C=0.5mol of CO2

and

1mol of O2=1mol of CO2, so

2.7mol of O2=2.7mol of CO2

now as we know that limiting reagent is that which produces least amount of the product and as the C is producing least amount of CO2 as compare to O2 therefore limiting reagent is C and as we determine the amount of product formed by the amount of limiting reagent so the amount of CO2 is 0.5mol

if required in mass then

mass of CO2=moles multiplied by molar mass

mass of CO2=0.5 multiplied by 48

mass of CO2=24 grams----answer

:)....hope you will be satisfied by the answer....:)