Answer:
[tex]n_{Na_2CO_3}=0.28molNa_2CO_3[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]Na_2CO_3(aq)+CaCl_2(aq)\rightarrow CaCO_3(aq)+2NaCl(aq)[/tex]
Hence, given the solution of calcium chloride, we can compute its reacting moles:
[tex]n_{CaCl_2}=0.35\frac{mol}{L}*803.1mL*\frac{1L}{1000mL}= 0.28molCaCl_2[/tex]
Thus, by knowing there is a 1:1 molar ratio between sodium carbonate and calcium chloride, we can easily compute the moles of sodium carbonate needed for a complete precipitation as shown below:
[tex]n_{Na_2CO_3}=0.28molCaCl_2*\frac{1molNa_2CO_3}{1molCaCl_2} \\\\n_{Na_2CO_3}=0.28molNa_2CO_3[/tex]
Best regards.
