Respuesta :
Answer:
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
