An excess of O2 reacted with 4.76 g of V. What is the theoretical yield of V2O5?

Question

An excess of O2 reacted with 4.76 g of V. What is the theoretical yield of V2O5?

______ g

How many grams of V2O5 were actually isolated if a 91.8% yield was obtained?

______ g

What volume of H2 could be produced at a temperature of 17 oC and a pressure of 755 torr by the reaction of 2.15 g of aluminum with hydrochloric acid?

_______ L H2

Answer 1

(i) 4V + 5O2(i.e.5molecules of oxygen)--->2V2O5

        (excess)

(4X50.9)                                              (50.9X4+16X10)  

=203.6                                               =363.6

4.76gm(given)                                          ?

Hence,the amount of V2O5=363.6/203.6 X 4.76

                                      =8.50gm(ans)

(ii) The amount of V2O5 when yield is 91.8%=91.8/100 X 8.50

                                                           =7.803 gm(ans)

(iii) 2Al + 6 HCl -----> 2AlCl3 + 3H2

    (27 x2)                              (3 X 22.4 lt.)

    =54                                       =67.2 lt.

   2.15gm                                  ?

  

Hence vol. of H2 produced =67.2/54 X 2.15

                                       =2.67 lt.

Now,P1 X V1/T1 =P2 X  V2 / T2

      755 X V1/ (17+ 273) = 760 X 22.4 / 273

      V1 =23.9 lt.(ans)