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?

2 Answers
Mar 29, 2017

Answer:

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

Mar 29, 2017

Answer:

#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 #