Question

15g of copper(ii) chloride reacts with 20g of sodium nitrate. Calculate the mass of sodium chloride formed?

I calculated the answer to be around 11.7g. Can someone help verify this?

Answer 1

12.15 grams

Explanation:

I got elemental weights as (Na=22.99 grams; Cu=63.55 grams; Cl=31.45 grams; N=14.01 grams; O=16 grams). The answer (Mass of sodium chloride after reaction) is 12.15 grams

Answer 2

`NaCl` formed = `13.03 g`

Explanation:

`CuCl_2` + `2NaNO_3` ------------> `2NaCl` + `Cu(NO_3)_2`

no. of moles of `CuCl_2` = `15g`/`134.45` g.mol -1 = `0.1115` moles ..............(1)

no. of moles of `NaNO_3` = `20g`/`84.99` g.mol -1 = `0.2353` moles........(2)

According to equation
1 mole of `CuCl_2` produces `NaCl` = `2` moles
so, `0.1115` moles of `CuCl_2` will produce `NaCl` = `2` x `0.1115` moles = `0.223` moles ........(3)

2 mole of `NaNO_3` produces `NaCl` = `2` moles
so, `0.2353` moles of `NaNO_3` will produce `NaCl` = `2` x `0.2353` moles...........(4)

`0.1115` moles of `CuCl_2` produces lesser number of moles of `NaCl`. hence `CuCl_2` is limiting reagent..................(5)

Therefore mass of `NaCl` produced = `0.223` moles.......(6)
Molar mass of `NaCl` = `58.44` g/mol
mass of `NaCl` = `0.223` moles x `58.44` g/mol = `**13.03** g`