Question

If 1. 3618 moles of AsF3 are allowed to react with 1. 0000 mole of C2Cl6, what would be the theoretical yield of AsCl3, in moles

Answer 1

Answer:

AsF3:C2CI6

4:3

1.3618 moles: 1.02135 moles(1.3618÷4×3)

C2CI6 is the limting reagent

So the number of moles for AsCI3 is 0.817 moles( number of moles of the limting reagant) ÷3 ×4 (according to ratio by balancing chemical equation)=1.09 moles(3 s.f.)

or

Balanced equation

4AsF3 + 3C2Cl6 → 4AsCl3 + 3C2Cl2F4

Use stoichiometry to calculate the moles of AsCl3 that can be produced by each reactant.

Multiply the moles of each reactant by the mole ratio between it and AsCl3 in the balanced equation, so that the moles of the reactant cancel, leaving moles of AsCl3.

Explanation: