Question #669e0

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Question #669e0

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Answer 1 · 2017-02-20T07:21:13

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

Explanation:

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

Given Conentration of Nitric acid = $16$ $16$ mol/L
Volume = $82 m l$ $82 ml$ = $0.082 L$ $0.082L$
Molarity = $\frac{n}{V}$ $n/V$
where n = no. of moles of $H N O 3$ $HNO3$ and $V$ $V$ = volume in $L$ $L$
n = $\frac{16}{0.082}$ $16/0.082$ = $1.312$ $1.312$ moles .................(2)

Chemical equation
$C u$ $Cu$ + $4 H N O_{3}$ $4HNO_3$ --> $C u {(N O_{3})}_{2}$ $Cu(NO_3)_2$ + $2 H_{2} O$ $2H_2O$ + $2 N O_{2}$ $2NO_2$

According to equation
1 mole of $C u$ $Cu$ produces 2 moles of $2 N O_{2}$ $2NO_2$
so, $0.297$ $0.297$ moles of $C u$ $Cu$ will produce $2 N O_{2}$ $2NO_2$ = $2$ $2$ X $0.297$ $0.297$ = $0.594$ $0.594$ moles...................(3)

In equation
$4$ $4$ moles of $4 H N O_{3}$ $4HNO_3$ produces 2 moles of $2 N O_{2}$ $2NO_2$
so, $1.312$ $1.312$ moles of $4 H N O_{3}$ $4HNO_3$ will produce $2 N O_{2}$ $2NO_2$ = ( $2$ $2$ x $1.312$ $1.312$ )/ $4$ $4$ = $0.656$ $0.656$ moles ............(4)

As given moles of $C u$ $Cu$ produces lesser moles of $2 N O_{2}$ $2NO_2$ hence $C u$ $Cu$ is the limiting reagent

a. The limiting reagent in the reaction is Copper

b. As solved above
$0.297$ $0.297$ moles of $C u$ $Cu$ will produce $2 N O_{2}$ $2NO_2$ = $2$ $2$ X $0.297$ $0.297$ = $0.594$ $0.594$ moles

$2 N O_{2}$ $2NO_2$ produced = $0.594$ $0.594$ moles
Molar mass of $2 N O_{2}$ $2NO_2$ = $46$ $46$ g/mol
mass = $0.594$ $0.594$ moles X $46$ $46$ g/mol = $27.324 g$ $27.324g$

$2 N O_{2}$ $2NO_2$ = $27.324 g$ $27.324g$

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Question #669e0

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