Question

What is the maximum mass of ammonia that can be formed when 28.17 grams of nitrogen gas reacts with 10.62 grams of hydrogen gas according to the following equation? Please round your answer to the nearest 0.01 gram.

N2 + 3 H2 à 2 NH3

Answer 1

34.26 grams of NH3 will be formed

Step-by-step explanation:

Step 1: Data given

Mass of nitrogen gas (N2) = 28.17 grams

Molar mass of N2 = 28.00 g/mol

Mass of hydrogen gas (H2) = 10.62 grams

Molar mass of H2 = 2.02 g/mol

Step 2: The balanced equation

N2(g) + 3H2(g) → 2NH3(g)

Step 3: Calculate moles

Moles = mass / molar mass

Moles N2 = 28.17 grams / 28.00 g/mol

Moles N2 = 1.006 moles

Moles H2 =10.62 grams / 2.02 g/mol

Moles H2 = 5.27 moles

Step 4: Calculate the limiting reactant

For 1 mon N2 we need 3 moles H2 to produce 2 moles NH3

N2 is the limiting reactant. It will completely be consumed (1.006 moles). H2 is in excess. There will react 3*1.006 = 3.018 moles

There will remain 5.27 - 3.018 = 2.252 moles

Step 5: Calculate moles NH3

For 1 mon N2 we need 3 moles H2 to produce 2 moles NH3

For 1.006 moles N2 we'll have 2*1.006 = 2.012 moles NH3

Step 6: Calculate mass NH3

Mass NH3 = moles NH3 * molar mass NH3

Mass NH3 = 2.012 moles * 17.03 g/mol

Mass NH3 = 34.26 grams

38.13 grams of NH3 will be formed

Answer 2

34.21g of ammonia

Step-by-step explanation:

N2 + 3 H2 ------->2 NH3

First find the limiting reactant

Using hydrogen

10.62 g of H2× 1.0 mole of H2/2g H2 ×2.0 mol of NH3/3.0 mol of H2× 17.0 g of NH3/1.0 mol of NH3= 60.18g of NH3

Using nitrogen

28.17 g of N2× 1.0 mole of N2/28g N2 ×2.0 mol of NH3/1.0 mol of N2 × 17.0 g of NH3/1.0 mol of NH3= 17.1g of NH3

Hence nitrogen gas is the limiting reactant.

From the reaction equation:

28.0 g of nitrogen gas gives 34.0g of ammonia

28.17 g of nitrogen gas gives 28.17×34.0/28.0= 34.21g of ammonia