Question

If the student uses 5.00 grams of (CH3)3CCO2H, how many grams of Mo(CO)6 will the student need to react completely with the (CH3)3CCO2H.

Answer 1

Answer – We are given the, mass of (CH₃)₃CCO₂H = 5.00 g

So balanced reaction between Mo(CO)₆ and (CH₃)₃CCO₂H is as follow

2 Mo(CO)₆ + 4 (CH₃)₃CCO₂H ---> Mo₂((CH₃)₃CCO₂)₄ + 12CO + 2H₂

First we need to calculate the moles of (CH₃)₃CCO₂H

Moles of (CH₃)₃CCO₂H =5.00 g / 102.132 g.mol^-1

                                        = 0.0490 moles

Now we need to calculate the moles of Mo(CO)₆ using the mole ratio, so

From the balanced reaction

4 moles of (CH₃)₃CCO₂H   = 2 moles of Mo(CO)₆

So, 0.0490 moles of (CH₃)₃CCO₂H   = ?

= 0.0245 moles of Mo(CO)₆

So, mass of Mo(CO)₆ = 0.0245 moles * 264.021 g/mol

                                    = 6.47 g

So, 6.47 grams of Mo(CO)₆ will the student need to react completely with the (CH3)₃CCO₂H.