how many moles of PCl5 can be produced from 50.0 g of Cl2
P4+ 10Cl2 → 4PCl5
M(Cl2)= 70.90 g/mol
M(PCl5)=208.24 g/mol
"n = \\frac{m}{M} \\\\"
n(Cl2) "= \\frac{50.0}{70.90}=0.705 \\;mol"
According to the reaction:
n(PCl5) "= \\frac{4}{10}n(Cl_2) = \\frac{4}{10} \\times 0.705 = 0.282 \\;mol"
m(PCl5) "= 0.282 \\times 208.24 = 58.74 \\;g"
Answer: 58.74 g
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