3A 2B --> 5C If compound A has a molar mass of 159.7 g/mol and compound C has a molar mass of 57.6 g/mole, how many grams of compound C will be produced from 18.24 grams of compound A and excess compound B

By stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction) the following amounts of reagent and products participate in the reaction:

A: 3 moles
B: 2 moles
C: 5 moles

The excess reagent will be that which is not completely depleted during the reaction.

The amount of product obtained from the reaction will always depend on the amount of limiting reagent in the reaction. Then, being B the excess reagent and therefore A the limiting reagent and knowing that compound A has a molar mass of 159.7 g/mole and compound C has a molar mass of 57.6 g/mole, by stoichiometry the following mass amounts of A and C participate in the reaction:

A: 3 moles* 159.7 g/mole= 479.1 g
C: 5 moles* 57.6 g/mole= 288 g

Then it is possible to apply the following rule of three: if by stoichiometry of the reaction 479.1 grams of A produce 288 grams of C, 18.24 grams of A, how much mass of C does it produce?

[tex]mass of C=\frac{18.24 grams of A*288 grams of C}{479.1 grams of A}[/tex]

mass of C= 10.96 grams

10.96 grams of compound C will be produced from 18.24 grams of compound A and excess compound B.