# What is the limiting reactant when 6.00 g of BaCl2 reacts with 5.00 g of Na3PO4 to form the precipitate Ba3(PO4)2?

May 7, 2017

$N {a}_{3} P {O}_{4}$

#### Explanation:

To find out the limiting reactant, you calculate the number of moles of the reactant and the limiting reactant is the one with the least number of moles.
The reaction is;
$3 B a C {l}_{2} + 2 N {a}_{3} P {O}_{4} \to B {a}_{3} {\left(P {O}_{4}\right)}_{2} + 6 N a C l$

No. of moles = $\frac{m a s s}{M r}$
In this case,

No. of moles ( $N {a}_{3} P {O}_{4}$) = $\frac{5.00}{601.93}$= 0.00831 moles
No. of moles of ($B a C {l}_{2}$) = $\left(\frac{6.00}{208.23}\right)$=0.0288 moles
This number of moles if for 1 molecule of the reactants.

But in this case 3 molecules of $B a C {l}_{2}$ reacts with 2 molecules of $N {a}_{3} P {O}_{4}$,

No. of moles of 3 molecules of $B a C {l}_{2}$= $0.0288 \cdot 3$=0.0864moles.
No. of moles of 2 molecules of $N {a}_{3} P {O}_{4}$=$0.00831 \cdot 2$=0.0166 moles.

So, as $N {a}_{3} P {O}_{4}$ has less number of moles than $B a C {l}_{2}$, $N {a}_{3} P {O}_{4}$ acts as the limiting reactant.

May 7, 2017

$B a C {l}_{2}$

#### Explanation:

you have to balance the equation
$3 B a C {l}_{2} + 2 N {a}_{3} P {O}_{4} = B {a}_{3} {\left(P {O}_{4}\right)}_{2} + 6 N a C l$
from the ratio of reaction you see that are need 3 mol of $B a C {l}_{2}$ for 2 mol of$N {a}_{3} P {O}_{4}$ to give one mol of$B {a}_{3} {\left(P {O}_{4}\right)}_{2}$
Then you have to calculate the mol of the reactants
$m o l = \frac{M g}{M M}$
$m o l B a C {l}_{2} = 6 , \frac{00}{208} , 23 = 0 , 0288 m o l$
$m o l N {a}_{3} P {O}_{4} = 5 , \frac{00}{163} , 94 = 0 , 0305 m o l$
You have less $B a C {l}_{2}$. This is the limitant reactant. It can react only 2/3 0,0288 of the mol of $N {a}_{3} P {O}_{4}$ =0,0192 mol and you obtain 1/3 0,0288 mol of$B {a}_{3} {\left(P {O}_{4}\right)}_{2}$ =0,0960