Question

Question #9c275

Answer 1

You can produce 10.1 g of `"Na"_3"PO"_4`.

Explanation:

The balanced equation is

`"3NaOH + H"_3"PO"_4 → "Na"_3"PO"_4 + 3"H"_2"O"`

`"Moles of NaOH"`

`= 1.11 × 10^23 color(red)(cancel(color(black)("formula units NaOH"))) × "1 mol NaOH"/(6.022 × 10^23 color(red)(cancel(color(black)("formula units NaOH")))) = "0.1843 mol NaOH"`

`"Moles of Na"_3"PO"_4 = 0.1842 color(red)(cancel(color(black)("mol NaOH"))) × ("1 mol Na"_3"PO"_4)/(3 color(red)(cancel(color(black)("mol NaOH")))) = "0.061 44 mol Na"_3"PO"_4`

`"Mass of Na"_3"PO"_4 = "0.061 44" "mol Na"_3"PO"_4 × ("163.94 g Na"_3"PO"_4)/(1 "mol Na"_3"PO"_4) = "10.1 g Na"_3"PO"_4`