What mass of PCl5 will be produced from the given masses of both reactants?
Calculate the number of moles of PCl5 that can be produced from 23.0 g of P4 (and excess Cl2).
| 0.743 mole |
Calculate the number of moles of PCl5 that can be produced from 53.0 g of Cl2 (and excess P4).
| 0.299 mole |
moles of P4 = 23/ 123.9 = 0.186
P4 + 10 Cl2 ---------------> 4 PCl5
1 4
0.186 ?
moles of PCl5 formed = 0.186 x 4 / 1
= 0.743
moles of PCl5 formed = 0.743
Cl2 moles = 53 / 70.91 = 0.747
P4 + 10 Cl2 ---------------> 4 PCl5
10 4
0.747 ?
moles of PCl5 formed = 0.747 x 4 / 10
= 0.299
moles of PCl5 formed = 0.299
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What mass of PCl5 will be produced from the given masses of both reactants?
What mass of PCL5 will be produced from the given masses of both reactants
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