If 25.0 g of nh3 and 38.9g of o2 react in the following reaction, how
many grams of no will be formed?
4 nh3 (g) + 5 o2 (g) → 4 no (g) + 6 h20 (g)

There will be formed 36.49 grams of NO

Explanation:

Step 1: The balanced equation

4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H20 (g)

Step 2: Data given

Mass of NH3 = 25.0 grams

Mass of O2 = 38.9 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of 02 = 32 g/mol

Molar mass of NO = 30.01 g/mol

Step 3: Calculate number of moles

Moles of NH3 = 25.0 grams / 17.03 g/mol

Moles of NH3 = 1.47 moles

Moles of O2 = 38.9 grams / 32 g/mol

Moles of O2 = 1.216 moles

Step 4: Calculate limiting reactant

For 4 moles of NH3 consumed, we need 5 moles of O2 to produce 4 moles of NO and 6 moles of H2O

O2 is the limiting reactant. It will complete be consumed (1.216 moles).

NH3 is the limiting reactant. There will 4/5 * 1.216 = 0.9728 moles of NH3 consumed.

There will remain 0.4972 moles of NH3.

Step 5: Calculates moles of NO

If there is consumed 4 moles of NH3, there is produced 4 moles of NO

For 1.216 moles of NH3 consumed, we will have 1.216 moles of NO

Step 6: Calculate mass of NO

Mass NO = moles NO * Molar mass NO

Mass NO = 1.216 moles * 30.01 g/mol = 36.49 grams

There will be formed 36.49 grams of NO

c, liquids evaporate, creating a gas