18.74 g of P4 is reacted with 9.6 g of Cl2 balanced is P4 (s) + 6 Cl2 (g) → 4 PCl3 (l) how many PlC3 can I make?
1 answer:
P4(s) + 6Cl2(g) → 4PCl3(l)
n P4 = m P4 / Mr P4
n P4 = 18.74 / 123.88
n P4 = 0.151 mol
n Cl2 = m Cl2 / Mr Cl2
n Cl2 = 9.6 / 70.9
n Cl2 = 0.135 mol
Cl2 act as <u>limiting reactant</u>
n PCl3 = (coef. PCl3)/(coef. Cl2) • n Cl2
n PCl3 = (4/6) • 0.135
n PCl3 = 0.09 mol
m PCl3 = n PCl3 • Mr PCl3
m PCl3 = 0.09 • 137.33
m PCl3 = 12.36 gr
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