You use 86.3 g #NO #and 25.6 g #H_2# in an experiment and produce 49.0 g of ammonia. How much excess reactant was used to make the 49.0 g of ammonia?

1 Answer
Jun 10, 2017

Answer:

Approx. #5*1/2*mol# dihydrogen gas.........

Explanation:

We need (i) a stoichiometric equation.....

#NO(g) + 5/2H_2(g) rarr NH_3(g)+H_2O(g)#

Which (I think) is balanced, as indeed it must be if we purport to represent chemical reality.

And then (ii) we work out equivalent quantities of product and reactant.

#"Moles of NO"=(86.3*g)/(30.01*g*mol^-1)=2.88*mol#

#"Moles of dihydrogen"=(25.6*g)/(2.02*g*mol^-1)=12.67*mol#

#"Moles of ammonia"=(49.0*g)/(17.03*g*mol^-1)=2.88*mol#

So you have got #100%# yield. Call the newspapers!

The excess dihydrogen is simply #{12.67-7.20}*mol=?*mol# where, by the given stoichiometry, #7.2*mol# of dihydrogen gas actually reacted.

Capisce?