Question

Given the equation NH3 + HCl →NH4Cl, find the number of particles of NH4Cl formed if 30.0 g of NH3 reacts with 10.0 g of

HCl.

Answer 1

n = m / M

n(NH₃) = 30.0 / 17.0 = 1.765 mol

n(HCl) = 10.0 / 36.5 = 0.274 mol

NH₃ is in excess, so calculate using n(HCl):

N(NH₄Cl) = N(HCl) = n(HCl) × N_A = 0.274 × 6.022 × 10²³ = 1.650 × 10²³

Answer: N(NH₄Cl) = 1.650 × 10²³.