Question

In: Chemistry

Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation:...

Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation: N2(g) + 3H2(g) → 2NH3(g)

1. What is the maximum mass of ammonia that can be produced from a mixture of 216.5 g of N2 and 51.40 g of H2? g

2. Which element would be completely consumed? (enter nitrogen or hydrogen)

3. What mass of the starting material would remain unreacted?

Solutions

Expert Solution

According to reaction 1 mole of nitrogen react with 3 mole of hydrogen to produce 2 mole of ammonia

molar mass of N2 =28.0134 g/mol and

molar mass of H2 = 2.01 g/mol then molar mass of 3H2 =6.03 g/mol

28.0134 gm of N2 react with 6.03 gm of H2 then 216.5 gm N2 require

216.56.03/28.0134 =46.60 gm of H2

N2 react completly and 51.40 - 46.60 = 4.8 gm H2 remain unreacted

molar mass of NH3 = 17.031 gm/mol then 2NH3 = 34.062 gm

1 mole of nitrogen react with 3 mole of hydrogen to produce 2 mole of ammonia and N2 react comletly that mean

28.0134 gm N2 produce 34.062 gm NH3 then 216.5 gm produce how many gm NH3 calculation are as follow

gm of NH3 = 216.534.062/28.0134 = 263.246 gm

maximum mass of ammonia = 263.246 gm

2) N2 consumed completely

3)

molar mass of N2 =28.0134 g/mol and

molar mass of H2 = 2.01 g/mol then molar mass of 3H2 =6.03 g/mol

28.0134 gm of N2 react with 6.03 gm of H2 then 216.5 gm N2 require

216.56.03/28.0134 =46.60 gm of H2

N2 react completly and 51.40 - 46.60 = 4.8 gm H2 remain unreacted


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