Question

molecules of i2 are produced by the reaction of 0.4321 mol of cucl2 according to the following equation. 2 cucl2 4 ki → 2 cui 4 kcl i2

Answer 1

$\: 1.237 \times 10^{23} molecules \: of \:I _{2} \: are$

$are\: produced \: and \: 53.74 \: g \: of \: I _{2} \: are$

produced from the reaction.

The overall balanced equation for the reaction is,

$2CuCl_{2} + KI→2CuI + 4KCl + I_{2}$

$Number \: of \: moles \: ofCuCl_{2}=0.4235 \: mol$

$2 \: moles \: of \: CuCl_{2} \: produce \: \: 1 \: mole \:$

$of \: I_{2}.$

$2 \: mole \: of \: CuCl_{2} = 1 \: mole \: of \: I_{2}$

$1 \: mole \: of \: I_{2} \: = 6.022 \times 10 ^{23} \: molecules \: of I_{2}$

So, the number of molecules produced in the reaction of

$I_{2} \: are$

$= \frac{0.4235 \times 6.022 \times 10 ^{23} }{2 }$

$= 1.237 \times 10^{23} \: molecules \:$

$1.237 \times 10^{23} \: molecules \: of \: I_{2} \: are \:$

produced from the reaction.

$Mass \: of \: I _{2} \: produced \: are \: is ,$

$= \frac{0.4235 \times 253.8}{2}$

$= 53.74 \: g$

$53.74 \: g \: of \: I _{2} \: produced \: in \: the \:$

reaction.

$Therefore , \: 1.237 \times 10^{23} molecules$

$of \: I_{2} \: are\: produced \: and \: 53.74 \: g \: of$

are produced in the reaction.

To know more about moles, refer to the below link:

brainly.com/question/26416088

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