Question

Please help!!

How many moles of sodium carbonate (Na2CO3) are required to precipitate the calcium ion from 803.1 mL of a 0.35 M CaCl2 solution?

Answer 1

`n_(Na_2CO_3)=0.28molNa_2CO_3`

Step-by-step explanation:

Hello,

In this case, the undergoing chemical reaction is:

`Na_2CO_3(aq)+CaCl_2(aq)\rightarrow CaCO_3(aq)+2NaCl(aq)`

Hence, given the solution of calcium chloride, we can compute its reacting moles:

`n_(CaCl_2)=0.35(mol)/(L)*803.1mL*(1L)/(1000mL)= 0.28molCaCl_2`

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:

`n_(Na_2CO_3)=0.28molCaCl_2*(1molNa_2CO_3)/(1molCaCl_2) \\\\n_(Na_2CO_3)=0.28molNa_2CO_3`

Best regards.