I think the answer might be false. I hope this helps!
Answer:
Na₃PO₄ + 3HCl —> 3NaCl + H₃PO₄
The coefficients are: 1, 3, 3, 1
Explanation:
_Na₃PO₄ + __HCl —> __NaCl + _H₃PO₄
The above equation can be balance as follow:
Na₃PO₄ + HCl —> NaCl + H₃PO₄
There are 3 atoms of Na on the left side and 1 atom on the right side. It can be balance by writing 3 before NaCl as shown below:
Na₃PO₄ + HCl —> 3NaCl + H₃PO₄
There are 3 atoms of Cl on the right side and 1 atom on the left side. It can be balance by writing 3 before HCl as shown below:
Na₃PO₄ + 3HCl —> 3NaCl + H₃PO₄
Now, the equation is balanced.
The coefficients are: 1, 3, 3, 1
Explanation:
(a) Formula that shows relation between
and
is as follows.
Here,
= 1
Putting the given values into the above formula as follows.
= 
= 
= 0.01316
(b) As the given reaction equation is as follows.

As there is only one gas so
,
= 1.20
Therefore, pressure of
in the container is 1.20.
(c) Now, expression for
for the given reaction equation is as follows.
![K_{c} = \frac{[CaO][CO_{2}]}{[CaCO_{3}]}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%7B%5BCaO%5D%5BCO_%7B2%7D%5D%7D%7B%5BCaCO_%7B3%7D%5D%7D)
=
= \frac{x^{2}}{(a - x)}[/tex]
where, a = initial conc. of 
=
= 0.023 M
0.0131 =
x = 0.017
Therefore, calculate the percentage of calcium carbonate remained as follows.
% of
remained =
= 75.46%
Thus, the percentage of calcium carbonate remained is 75.46%.