How many moles of `O` are present in `4.9 g` of `H_(3) PO_(4)`? (Atomic weight of `P, O` and `H = 31, 16, 1`)

Explanation

Molecular weight of `H_(3) PO_(4) = 1 xx 3 + 31 + 16 xx 4 = 98 g`
`98 g = 1 "mole of" H_(3) PO_(4) = 4` mole of `O`
`4.9 g of H_(3) PO_(4) = (4)/(98) xx 4.9 = 0.2` mole of O