Question

Given the balanced reaction: N_{2 (g)} + 3H_{2 (g)} 

⟶

 2NH_{3 (g)}

_{If you had 15 grams of nitrogen gas mixed with 4 grams of hydrogen, what would the maximum amount (in grams) of ammonia that would be theoretically produced?}

Answer 1

I. Calculate moles of substances involved

Moles of Nitrogen available;

$Moles = \dfrac{mass}{Molar mass}$

$Moles = \dfrac{15g}{14g/mol} = 1.07mol$

Moles of hydrogen available;

$Moles = \dfrac{4g}{4g/mol} = 4 moles$

Moles of hydrogen that would react with 1.07mol of Nitrogen;

mole ration of N2: H2 = 1:3

$Moles of hydrogen= \dfrac{3}{1} x 1.07mol = 3.2mol$

But there are 4 moles of hydrogen, thus, nitrogen is the limiting reagent.

Moles of ammonia that would be produced by 1.07moles of nitrogen;

Mole ratio of N2: NH3 = 1:2

$Moles of ammonia =\dfrac{2}{1}x1.07mol = 2.14mol$

Mass of ammonia = moles x RFM ; [RFM = 14 + (1x3) = 17]

= 2.14 mol x 17g/mol

= 36.38g