In a particular experiment , 272 g of phosphorus, `P_(4)` , reacted with excess of oxygen of form `P_(4)O` in ` 89 . 5 % ` yield. In the second step of the reaction, a `97.8%` yield of `H_(3) PO_(4)` was obtained. What mass of `H_(3) PO_(4)` was obtained ?

Explanation

Applying POAC for P atoms for the following steps. In the first step, for `89.5%` yield,
moles of `P_(4) O_(10)` produced = moles of `P_(4) xx 0.895`
`= (272)/(124) xx 0.895 = 1.9632`
In the second step, for `97.8% ` yield,
moles of `H_(3) PO_(4)` produced ` = 4 xx ` moles of `P_(4) O_(10) xx 0.978`
`= 4 xx 1.9632 xx 0.978 = 7.680`
`:.` wt. of `H_(3) PO_(4)` produced = `7 . 680 xx 98 g`
= 752 . 65 g