# Chemical reaction calculations

Hi

I got a couple of questions regarding the following chemical reaction.

First some calculations:

$$90,0 \textrm{mL of} \ 0,250 \ \textrm{M} \ BaCl_{2}$$ is mixed with $$0,500 \textrm{M}$$ of $$Na_{2} CO_{3}$$ which generats residuum.

a) $$BaCl_{2} \rightarrow Ba^{2+} + 2Cl_{2}^{-} \ (i)$$

$$Na_{2} CO_{3} \rightarrow 2Na^{+} + CO_{3}^{2-} \ (ii)$$

b) The mol mass of (i) and (ii) are:

$$C(BaCl_{2}) = \frac{n(BaCl{2})}{V(BaCl{2})} \rightarrow n(BaCl_{2}) = C(BaCl_{2})} \cdot V(BaCl_{2}) = 0,09 L \cdot 0,25 L/mol = 0,0,225 \textrm{mol}$$

$$C(Na_{2}CO_{3}) = \frac{n(Na_{2}CO_{3})}{V(Na_{2}CO_{3}} \rightarrow n(Na_{2}CO_{3}) = C(Na_{2}CO_{3}) \cdot V(Na_{2}CO_{3}) = 0,08 L \cdot 0,50 L/mol = 0,040 \textrm{mol}$$

c) The Chemical reaktion schema for the mixture of $$BaCl_{2}$$ and $$Na_{2} CO_{3}$$ is:

$$Ba Cl_{2} + Na_{2} CO_{3} \rightarrow BaCO_{3} + 2NaCl$$

I got two questions:

1) I would like to calculate the residuum generated in $$Ba Cl_{2} + Na_{2} CO_{3} \rightarrow BaCO_{3} + 2NaCl \ ???$$

2) Second I would like to Calculate $$[Na^{+}]$$ in c) ??

Sincerely
Fred
Denmark

chem_tr
When the mole ratios of both barium chloride and sodium carbonate are equal, then the precipitation will be finished. So 0,225 moles of $BaCl_2$ will need another 0,225 moles of $BaCO_3$. The "residuum" you mention is probably the precipitate, i.e., $BaCO_3$, so the mole number is the same as $BaCl_2$. Just calculate the molar mass for barium carbonate, and multiply with the mole number to learn the mass of it.
About second question, we first need to assume that any solid precipitate does not change the total volume of the medium. Then you can find the total liquid volume, as two solutions are mixed. The last action will be to find the new concentration of $[Na^+]$, from mole number and the new volume. It is easy, I think.