## Please help with lab

**Anonymous**

Hi can anyone help me with this problem

Suppose you wanted to add just enough 0.40 molary of calcium chloride solution to 75.0 mL of 0.60 molary of sodium carbonate solution for all the sodium carbonate and calcium chloride to react.

a. what volume of 0.40 molary of calcium chloride solution would be required?

b. what mass of calcium carbonate would be produced (assuming a 100% yield)?

WebElements| 14 January 2004 - 11:31am## Re: Please help with lab

[quote="rsx"]Hi can anyone help me with this problem

Suppose you wanted to add just enough 0.40 molary of calcium chloride solution to 75.0 mL of 0.60 molary of sodium carbonate solution for all the sodium carbonate and calcium chloride to react.

a. what volume of 0.40 molary of calcium chloride solution would be required?

b. what mass of calcium carbonate would be produced (assuming a 100% yield)?[/quote]

The number of moles of sodium carbonate used is = 0.075 * 0.6

As the reaction is 1:1, the required amount of calcium chloride is X * 0.4 where X is the volume in litres.

So 0.075 * 0.6 = X * 0.4 - solve for X

The weight of sodiuim carbonate is 0.075 * 0.6 * mol wt of sodium carbonate. Make sure to get your units correct.