# How would we write the equation between Na_3PO_4 and Pb(NO_3)_2, given 17.3*g of the former, and 12.4*g of the latter?

Apr 5, 2016

$2 N {a}_{3} P {O}_{4} \left(a q\right) + 3 P b {\left(N {O}_{3}\right)}_{2} \left(a q\right) \rightarrow P {b}_{3} {\left(P {O}_{4}\right)}_{2} \left(s\right) \downarrow + 6 N a N {O}_{3} \left(a q\right)$

#### Explanation:

$\text{Moles of lead nitrate}$ $=$ $\frac{12.4 \cdot \cancel{g}}{331.2 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $=$ $0.0374 \cdot m o l$.

$\text{Moles of trisodium phosphate}$ $=$ $\frac{17.3 \cdot g}{163.94 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.106 \cdot m o l$.

There is a slight excess of lead nitrate: $3 \times 0.0374 \cdot m o l \text{ lead nitrate} = 0.112 \cdot m o l$. See if you can work the excess in grams.

Note that if we did the actual experiment, we would probably get $P b H P {O}_{4} \left(s\right)$... i.e. $\text{lead(II)biphosphate...}$