Question

Balance the chemical equation given below, and calculate the volume of nitrogen monoxide gas produced when 8.00 g of ammonia is reacted with 12.0 g of oxygen at 25°C? The density of nitrogen monoxide at 25°C is 1.23 g/L.___ NH3(g) + ___ O2(g) ? ___ NO(g) + ___ H2O(l)A) 4.88LB) 7.38LC) 11.5LD) 17.3LE) 19.2L

Answer 1

Answer:

The volume of NO is 7.32 L  (option B 7.28L is the closest to this)

Explanation:

Step 1: Data

Mass of ammonia = 8.00 grams

Mass of oxygen = 12.0 grams

Temperature = 25.0 °C

Density of NO = 1.23 g/L

Step 2: The balanced equation

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)

Step 3: Calculate moles ammonia

Moles ammonia = 8.00 grams / 17.03 g/mol

Moles ammonia = 0.470 moles

Step 4: Calculate moles oxygen

Moles O2 = 12.0 grams / 32.0 g/mol

Moles O2 = 0.375 moles

Step 5: Calculate limiting reactant

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

O2 is the limiting reactant. It will completely be consumed (0.375 moles).

NH3 is in excess. There react 4/5*0.375 moles = 0.300 moles

There remain 0.470 - 0.300 = 0.170 moles NH3

Step 6: Calculate moles NO

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

For 0.375 moles O2 we'll have 0.300 moles NO produced

Step 7: calculate mass NO

Mass NO = moles * molar mass

Mass NO = 0.300 moles * 30.01 g/mol

Mass NO = 9.003 grams

Step 8: Calculate volume NO

Volume NO = mass NO / density

Volume NO = 9.003 grams / 1.23 g/L

Volume NO = 7.32 L

The volume of NO is 7.32 L  (option B 7.28L is the closest to this)