Question

What mass of P4O10 will be obtained from the reaction of 1.33 g of P4 and 5.07 g of oxygen?

Answer 1

P4 + 5 O2 = 2 P2O5

Molar mass of P4 is 123.9 g / mol.
so we have (1.33g / 123.9 g/ mol) moles of P4 = 0.011 moles
according to the equation these will react with 0.011 * 5 moles of O2 = 0.054 moles
but we have 5.07 / 32 moles of O2 = 0.158 moles
which is >>> 0.054 moles
so the O2 is in excess and P4 is the limiting reagent
according to the equation 0.011 moles of P4 give 2 * 0.011 moles of P2O5 = 0.022 moles
Molar mass of P2O5 is 141.9 g/mol
so mass of P2O5 formed = (0.022 mol * 141.9 g/ mol) = 3.12 g

so P4 is the limiting reagent
and
3.12 g of P2O5 are formedP4 + 5 O2 = 2 P2O5
Molar mass of P4 is 123.9 g / mol
so we have (1.33g / 123.9 g/ mol) moles of P4 = 0.011 moles
according to the equation these will react with 0.011 * 5 moles of O2 = 0.054 moles
but we have 5.07 / 32 moles of O2 = 0.158 moles
which is >>> 0.054 moles
so the O2 is in excess and P4 is the limiting reagent
according to the equation 0.011 moles of P4 give 2 * 0.011 moles of P2O5 = 0.022 moles
Molar mass of P2O5 is 141.9 g/mol
so mass of P2O5 formed = (0.022 mol * 141.9 g/ mol) = 3.12 g

so P4 is the limiting reagent and 3.12 g of P2O5 are formed