Question

consider the following equation.
$pb(no_3)_2(aq)+k_3po_4(aq)\to pb_3(po_4)_2$
calculate the quantity, in moles, of $pb(no_3)_2$ that will be required to produce 365 grams of $pb_3(po_4)_2$.

Explanation:

Step1: Calculate moles of $Pb_3(PO_4)_2$

First, find the molar mass of $Pb_3(PO_4)_2$. The molar mass of $Pb$ is approximately $207.2\ g/mol$, $P$ is approximately $30.97\ g/mol$, and $O$ is approximately $16.00\ g/mol$.
The molar mass of $Pb_3(PO_4)_2=3\times207.2 + 2\times(30.97+4\times16.00)= 3\times207.2+2\times(30.97 + 64.00)=621.6+2\times94.97=621.6 + 189.94 = 811.54\ g/mol$.
The number of moles of $Pb_3(PO_4)_2$, $n=\frac{m}{M}$, where $m = 365\ g$ and $M = 811.54\ g/mol$. So $n=\frac{365\ g}{811.54\ g/mol}\approx0.45\ mol$.

Step2: Use stoichiometry

The balanced chemical equation is $3Pb(NO_3)_2(aq)+2K_3PO_4(aq)
ightarrow Pb_3(PO_4)_2(s)+6KNO_3(aq)$.
The mole - ratio of $Pb(NO_3)_2$ to $Pb_3(PO_4)_2$ is $3:1$.
If the number of moles of $Pb_3(PO_4)_2$ is $n_{Pb_3(PO_4)_2}=0.45\ mol$, then the number of moles of $Pb(NO_3)_2$, $n_{Pb(NO_3)_2}=3\times n_{Pb_3(PO_4)_2}$.
$n_{Pb(NO_3)_2}=3\times0.45\ mol = 1.35\ mol$.

Answer:

$1.35\ mol$