The balanced equation for considered reaction can be written as follows
"PH_3 (g) + 4 Cl_2 (g) \\rightarrow PCl_5 (g) + 3 HCl (l)"
"\\frac{m(Cl_2)}{M(Cl_2) \\cdot n(Cl_2)} = \\frac{m(PCl_5)}{M(PCl_5) \\cdot n(PCl_5)}"
"m(Cl_2) = \\frac{m(PCl_5) \\cdot M(Cl_2) \\cdot n(Cl_2)}{M(PCl_5) \\cdot n(PCl_5)}"
"M(Cl_2) = 2 \\cdot 35.453 = 70.906 \\; g\/mol"
"M(PCl_5) = 1 \\cdot 30.974 + 5 \\cdot 35.453 = 208.239 \\; g\/mol"
"m(Cl_2) = \\frac{0.015 \\cdot 70.906 \\cdot 4}{208.239 \\cdot 1} \\approx 0.0204 \\; kg = 20.4 \\; g"
Answer: 20.4 g.
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