Question

please check if I did this right ...

calculate the mass of white solid CaCO3 that forms when 98ml of a 0.1M Ca(NO3)2 solution is mixed with 54 ml of a 2.5M Na2CO3 solution.

Ca(NO3)2 + 2NaCO3 -> Ca(CO3)2 + 2NaNO3

.098*0.1 =.0090
.054*2.5 =.135

limiting reactant is Ca(NO3)2

.0090 * 160.1 = 1.44g

Answer 1

the balanced equation should be this:

Ca(NO3)2+ Na2CO3>>CaCO3 + 2NaNO3

You had two formulas wrong.

Answer 2

Thank you

so with those corrections my new calculations would be

.0090*100.05 = .90045g

Hopefully this time it is right.

Obviously I am not chemisrty gifted.

Answer 3

so with those corrections my new calculations would be

.0090*100.05 = .90045g
<b>The 0.0090 number you have comes from 0.098 x 0.1 = 0.0098 and you made a typo when you entered it. Then 0.0098 x 100.05 = 0.98 g
</b>

Answer 4

i have a question of why did you put 2 after the parentheses Ca(CO3)2

isn't Ca has the charge number of 2+
and the CO3 has the charge of negative 2