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2 months ago

A mixture of KCl and KNO3

2 answers:

8 0

The percentage of KCl present in the mixture is approximately 40%If we consider a 100-gram sample of the mixture:ω(K) = 44.20% ÷ 100% = 0.442.
m(K) = 0.442 · 100 g = 44.2 g.n(K) = 44.2 g ÷ 39.1 g/mol.
n(K) = 1.13 mol.
n(KCl) + n(KNO₃) = n(K)
m(KCl) = x.
m(KNO₃) = y.
Two equations:
1) x + y = 100 g.
2) m(KCl)/M(KCl) + m(KNO₃)/M(KNO₃) = 1.13 mol.
x/74.55 g/mol + y/101.1 g/mol = 1.13 mol.
From the first equation, we find x = 100 - y and substitute into the second equation:
(100 - y)/74.55 + y/101.1 = 1.13 /×101.1.
135.61 - 1.356y + y = 114.24.
0.356y = 22.37 g.
y = 62.83 g.
Thus, x = m(KCl) = 100 g - 62.83 g = 37.17 g.

KiRa [2.9K]2 months ago

6 0

Answer:

The precise answer is option (a).

Explanation:

Assuming the total mass of the mixture is 100 g

let the mass of KCl be x

and the mass of

be y $KNO_3$

x + y = 100...(1)

Moles of KCl =

$\frac{x}{75.5 g/mol}$

Moles of K in KCl =

(1 mol is found in 1 mole of the compound) $\frac{x}{75.5 g/mol}$

Moles of

$KNO_3=\frac{y}{102 g/mol}$ in

(1 mol is contained in 1 mole of the compound) $KNO_3$ $\frac{y}{102 g/mol}$ The percentage of potassium in the mixture is 44.20 %

In a 100 g sample, the potassium mass = 44.20 g

Moles of potassium are

$\frac{44.20 g}{40 g/mol}=1.105 mol$ Thus,

..(2)

$\frac{x}{75.5 g/mol}+\frac{y}{102 g/mol}=1.105 mol$ By solving equations (1) and (2):

we obtain y = 63.78 g, x=36.22 g

Calculating the percentage of KCl:

$\frac{36.22 g}{100g}\times 100=36.22 g\%$

The percent of KCl in the mixture is approximately 40%. Therefore, option (a) is correct.

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