P4(s)+10Cl2(g)→4PCl5(g) What mass of PCl5 will be produced from the given masses of both reactants? Express your answer to three significant figures and include the appropriate units. 0.872 mole and 0.310 reactants
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0.872 moleWhat is the percent yield if the actual yield from this reaction is 347 g ? Express your answer to three significant figures and include the appropriate units. |
P4(s)+10Cl2(g)→4PCl5(g)
1 mole of P4 reacts with 10 moles of Cl2
M moles of P4 reacts with 0.870 moles of Cl2
M = ( 1x0.870) / 10
= 0.0870 moles of Cl2
So (0.310 - 0.0870 ) moles of P4 left unreacted which is excess reactant
Since all the mass of Cl2 completly reacted it is the limiting reactant.
10 moles of Cl2 produces 4 moles of PCl5
0.870 moles of Cl2 produces (4x0.870) / 10 = 0.348 moles of PCl5
So mass of PCl5 produced = number of moles x molar mass
= 0.348 mol x 208.5 g/mol
= 72.6 g ----> This is the theoretical yield
So percent yield = ( actual yield / theoretical yield) x 100
= ( 34.7 / 72.6) x 100
= 47.8 %
P4(s)+10Cl2(g)→4PCl5(g) What mass of PCl5 will be produced from the given masses of both reactants? Express...
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