# 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?

Mar 29, 2017

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

$N a C l$ formed = $13.03 g$

#### Explanation:

$C u C {l}_{2}$ + $2 N a N {O}_{3}$ ------------> $2 N a C l$ + $C u {\left(N {O}_{3}\right)}_{2}$

no. of moles of $C u C {l}_{2}$ = $15 g$/$134.45$ g.mol -1 = $0.1115$ moles ..............(1)

no. of moles of $N a N {O}_{3}$ = $20 g$/$84.99$ g.mol -1 = $0.2353$ moles........(2)

According to equation
1 mole of $C u C {l}_{2}$ produces $N a C l$ = $2$ moles
so, $0.1115$ moles of $C u C {l}_{2}$ will produce $N a C l$ = $2$ x $0.1115$ moles = $0.223$ moles ........(3)

2 mole of $N a N {O}_{3}$ produces $N a C l$ = $2$ moles
so, $0.2353$ moles of $N a N {O}_{3}$ will produce $N a C l$ = $2$ x $0.2353$ moles...........(4)

$0.1115$ moles of $C u C {l}_{2}$ produces lesser number of moles of $N a C l$. hence $C u C {l}_{2}$ is limiting reagent..................(5)

Therefore mass of $N a C l$ produced = $0.223$ moles.......(6)
Molar mass of $N a C l$ = $58.44$ g/mol
mass of $N a C l$ = $0.223$ moles x $58.44$ g/mol = $\ast 13.03 \ast g$