Answer to Question #73073 in General Chemistry for Tristan
2018-01-31T00:05:27-05:00
How many moles of PCl5 can be produced from 60.0 g of Cl2 (and excess P4)?
Express your answer to three significant figures and include the appropriate units.
1
2018-02-01T05:02:49-0500
2P + 5Cl2 => 2PCl5 m(Cl2 )= 60.0 g m(PCl5) - ? n(Cl2) = m(Cl2)/ M(Cl2) = 60.0 g / 71 g/mol = 0.845 mol 2n(PCl5) = 5n(Cl2 ) n(PCl5) = 2n(Cl2 )/ 5 = 2*0.845 mol/ 5 = 0.338 mol m (PCl5) = n(PCl5) *M(PCl5) = 0.338 mol*208 g/mol = 70.304 g
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