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1.81 g H2 is allowed to react with 10.2 g N2, producing 2.19 g NH3.What is the theoretical yield in grams for this reaction under the given conditions?
3H2(g)+N2(g)→2NH3(g)

Sagot :

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The theoretical yield : = 10.251 g

Further explanation

Given

Reaction

3H₂(g)+N₂(g)→2NH₃(g)

1.81 g H₂

10.2 g N₂

2.19 g NH₃

Required

The theoretical yield

Solution

Find limiting reactant :

H₂ : 1.81 g : 2 g/mol = 0.905 mol

N₂ : 10.2 g : 28 g/mol = 0.364 mol

mol : coefficient

H₂ = 0.905 : 3 = 0.302

N₂ = 0.364 : 1 = 0.364

H₂ as a limiting reactant(smaller ratio)

Moles NH₃ based on H₂, so mol NH₃ :

= 2/3 x mol H₂

= 2/3 x 0.905

=0.603

Mass NH₃ :

= mol x MW

=0.603 x 17 g/mol

= 10.251 g

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