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.
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) × NA = 0.274 × 6.022 × 1023 = 1.650 × 1023
Answer: N(NH4Cl) = 1.650 × 1023.
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