The number of moles of N2O produced when 26.5 g N2 reacts with excess oxygen?

a.  0.964 moles of N₂O will be produced

b. mass of N₂ required = 7.31 g of N₂

c. Number of particles of N₂= 5.87 * 10²² molecules of N₂

Explanation:

a. Balanced equation of the reaction is given below: 2N₂ + O₂ ---> 2N₂O

Molar mass of N₂ = 28 g/mol

molar mass of O₂ = 32 g/mol

molar mas of N₂O = 44g/mol

From the equation of reaction, 2 moles of nitrogen gas reacts with 1 mole of oxygen gas to produce 2 moles of N₂O

number of moles of N₂ present in 26.5 g = 26.5 g/28 g/mol = 0.946 moles

Since oxygen gas is in excess in the reaction, the limiting reactant is N₂

Mole ratio of N₂ to N₂O is 1 : 1

Therefore, 0.964 moles of N₂ will react with excess oxygen gas to produce 0.964 moles of N₂O

b. number of moles of N₂O in 11.5 g of N₂O = 11.5/44 = 0.261 moles

0.261 moles of N₂O will be produced by 0.261 moles of N₂

mass of N₂ present in 0.261 moles = 0.261 * 28

mass of N₂ required = 7.31 g of N₂

c. number of moles of O₂ present in 1.56 g of O₂ = 1.56/32 = 0.04875moles

Moles of N₂ required to react with 0.4875 moles of O₂ = 2 * 0.04875 = 0.0975 moles

Number of particles of N₂ present in 0.0975 moles = 6.02 * 10²³ * 0.0975

Number of particles = 5.87 * 10²² molecules of N₂