Question

if 0.650 solution of na2co3 will be use to precipitate ca+2 from solution, determin the amount of solution in grams needed to precipitate 5.00x102ml of 0.01120m ca+?(atomic mass:na=22.99 g/mol;c=12.01g/mol;0=16.00g/mol;ca=40.08g/mol;ans.91.3g

Answer 1

$Ca^{2+}+ Na_2CO_3 \rightarrow CaCO_3 + 2Na^+$

At 5.00x102ml of 0.0112 m solution is $5x102x0.0112/1000=0.005712 mole$ of $Ca^{2+}$

From reaction is that $n(Na_2CO_3)=n(Ca^{2+})=0.005712$

$M(Na_2CO_3)=22.99x2+12.01++16x3=105.99 g/mol$

So 0.005712 is $0.005712x105.99=0.6054 g$ of $Na_2CO_3$

As solution concentration is 0.65, then it needs 0.6054/0.65=0.9314 g of solutuon.