# Question #669e0

Feb 20, 2017

a. Limiting reagent = $C u$
b. mass of NO2 produced = $27.324 g$

#### Explanation:

Given mass of Cu = $18.9 g$
Molar mass of Cu $63.55$ g/mol
No. of moles of Cu = $\frac{18.9}{63.55}$ = $0.297$moles ............(1)

Given Conentration of Nitric acid = $16$ mol/L
Volume = $82 m l$ = $0.082 L$
Molarity = $\frac{n}{V}$
where n = no. of moles of $H N O 3$ and $V$= volume in $L$
n = $\frac{16}{0.082}$ = $1.312$ moles .................(2)

Chemical equation
$C u$ + $4 H N {O}_{3}$ --> $C u {\left(N {O}_{3}\right)}_{2}$ + $2 {H}_{2} O$ + $2 N {O}_{2}$

According to equation
1 mole of $C u$ produces 2 moles of $2 N {O}_{2}$
so, $0.297$ moles of $C u$ will produce $2 N {O}_{2}$ = $2$ X $0.297$ = $0.594$ moles...................(3)

In equation
$4$ moles of $4 H N {O}_{3}$ produces 2 moles of $2 N {O}_{2}$
so, $1.312$ moles of $4 H N {O}_{3}$ will produce $2 N {O}_{2}$ = ($2$ x $1.312$)/ $4$ = $0.656$ moles ............(4)

As given moles of $C u$ produces lesser moles of $2 N {O}_{2}$ hence $C u$ is the limiting reagent

a. The limiting reagent in the reaction is Copper

b. As solved above
$0.297$ moles of $C u$ will produce $2 N {O}_{2}$ = $2$ X $0.297$ = $0.594$ moles

$2 N {O}_{2}$ produced = $0.594$ moles
Molar mass of $2 N {O}_{2}$ = $46$ g/mol
mass = $0.594$ moles X $46$ g/mol = $27.324 g$

$2 N {O}_{2}$ = $27.324 g$