Question

g For the following reaction, 5.07 grams of nitrogen gas are allowed to react with 5.40 grams of hydrogen gas . nitrogen(g) + hydrogen(g) ammonia(g) What is the maximum mass of ammonia that can be formed? grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams

Answer 1

6.16 grams of NH3 will be produced. N2 is the limiting reactant. H2 is in excess, there will remain 4.30 grams of H2

Step-by-step explanation:

Step 1: Data given

Mass of nitrogen gas(N2) = 5.07 grams

Mass of hydrogen gas (H2)

Molar mass N2 = 28.0 g/mol

Molar mass 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 = 5.07 grams / 28.0 g/mol

Moles N2 = 0.181 moles

Moles H2 = 5.40 grams / 2.02 g/mol

Moles H2 = 2.67 moles

Step 4: Calculate the limiting reactant.

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

N2 is the limiting reactant. It will completely be consumed (0.181 moles). H2 is in excess. There will react 3*0.181 = 0.543 moles

There will remain 2.67 - 0.543 = 2.127 moles

Step 5: Calculate mass of H2 that remains

Mass H2 = 2.127 moles * 2.02 g/mol

Mass H2 = 4.30 grams

Step 6: Calculate moles ammonia

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

For 0.181 moles N2 we need 2*0.181 = 0.362 moles NH3

Step 7: Calculate mass NH3

Mass NH3 = moles NH3 * molar mass NH3

Mass NH3 = 0.362 moles * 17.01 g/mol

Mass NH3 = 6.16 grams

6.16 grams of NH3 will be produced. N2 is the limiting reactant. H2 is in excess, there will remain 4.30 grams of H2