• DATA GIVEN:

Data
Mass of NH3 = 3.25g
Mass of O2 = 3.5g
Equation
4NH3(g) + 502(g) 4NO(g) + 6H20(g)
• a/ The limiting reagent

Mole of NH3 = mass /Molar mass
= 3.25g /17g/mole
= 0.1911764 mole
Mole ratio for NH3 = Mole /Coefficient mole
= 0.1911764 mole /4 mole
= 0.047794
Mole for O2 = Mass /Molar mass
= 3.5g /32g/mole
= 0.109375
Mole ratio for O2 = Mole /Mole coefficient
= 0.109375 /5
= 0.021875
a/ The limiting reagent is oxygen gas (O2)
• b/ To calculate the mass of NO formed

To use limiting reagent to know the ammoniaformed from mole equivalent
5 moles of O2 Ξ 4 moles of NO
0.109375 mole of O2 Ξ ?? of NO

= 0.109375 × 4 /5
= 0.0875 moles
To convert 0.0875 moles of NO into mass
Mass = 0.0875 moles
= 2.625g
2.625g of NO was formed.
• c/ To find the mass of excess reaction remain from mole equivalent between NH3 and O2

4 moles of NH3 Ξ 5 mole O2
?? Ξ 0.109375 O2

4 moles × 0.109375 mole /5 moles
= 0.0875 moles of NH3
To convert mole into mass.
Mass = moles × molar mass
= 0.0875 mole × 17g/mole
= 1.4875g
Mass remain of NH3 = original mass - mass reacted
= 3.25g - 1.4875g
= 1.7625g
1.7625g of ammonia (NH3) was remained after reaction