Question

Calculate the volume in ml of 0.200 M na2co3 needed to produce 2.00 g of caco3 there is an excess of cacl2

Answer 1

Answer:

$V=100mL$

Explanation:

Hello.

In this case, since the chemical reaction is:

$Na_2CO_3+CaCl_2\rightarrow CaCO_3+2NaCl$

We next compute the moles of sodium carbonate from the 2.00 grams of calcium carbonate via their 1:1 mole ratio in the chemical reaction:

$n_{Na_2CO_3}=2.00gCaCO_3*\frac{1molCaCO_3}{100.09gCaCO_3}*\frac{1molNa_2CO_3}{1molCaCO_3} \\\\n_{Na_2CO_3}=0.0200molNa_2CO_3$

Thus, by knowing the molarity, we compute the volume:

$M=\frac{n}{V}\\ \\V=\frac{n}{M}=\frac{0.0200mol}{0.200mol/L}\\ \\V=0.100L*\frac{1000mL}{1L}\\ \\V=100mL$

Best regards.