Question

Consider the reaction of lead(II) nitrate reacting with sodium phosphate to create lead(II) phosphate and sodium nitrate. If 37.1 g of lead(II) phosphate was created in the reaction, how much (mass) sodium phosphate was originally needed to produce this amount of product

Answer 1

Mass of sodium phosphate needed : 7.54 g

Further explanation

Reaction

3Pb(NO₃)₂ + 2Na₃PO₄ → Pb₃(PO₄)₂ + 6NaNO₃

mol Pb₃(PO₄)₂(MW=811.54272 g/mol)

$\tt \dfrac{37.1}{811.54272}=0.046$

mol Na₃PO₄ = mol  Pb₃(PO₄)₂=0.046

mass Na₃PO₄(MW=163,94 g/mol) :

$\tt 0.046\times 163.94=7.54124~g$