Question

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?

Answer 1

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?