Question

Which reactant, and how much of it, remains after the reaction of 40.0g of V₂O₅ (FW=181.9g/mol) with 40.0g of calicuim based on the following chemical equation?

V₂O₅ + 5Ca----> 2V+5CaO

Possible Answers

0.810g Ca

3.6g Ca

.801g V₂O₅

3.6g V₂O₅

4.52g V₂O₅

Answer 1

Answer – We are given, mass of V₂O₅ = 40.0 g , mass of Ca = 40.0 g

Reaction –

V₂O₅ + 5Ca----> 2V+5CaO

First we need to determine the limiting reactant

For that we need to calculate the moles of each reactant first

Moles of V₂O₅ = 40.0 g / 181.9 g.mol^-1

                         = 0.220 moles

Moles of Ca = 40.0 g / 40.078 g.mol^-1

                     = 0.998 moles

Now we need to calculate moles of any one product from both reactant

From the balanced equation

1 moles of V₂O₅ = 2 moles of V

So, 0.220 moles of V₂O₅ = ?

= 0.440 moles of V

Moles of V from Ca

5 moles of Ca = 2 moles of V

So, 0.998 moles of Ca = ?

= 0.400 moles of V

So we got the lowest moles of V from the moles of Ca, so limiting reactant is Ca and the excess reactant is V₂O₅.

Moles of V₂O₅ used –

5 moles of Ca = 1 moles of V₂O₅

So, 0.998 moles of Ca = ?

= 0.200 moles

So moles of V₂O₅ remaining = 0.220 – 0.200

                                               = 0.020 moles

So , mas of V₂O₅ = 0.020 moles * 181.9 g/mol

                           = 3.69 g

So answer is 3.6g V₂O₅