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?

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Answer 1 · 2017-06-10T06:03:24

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

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?

You use 86.3 g NO NO and 25.6 g H_2H_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?