Question

A mixture of potassium bromide and potassium hydroxide has a total mass of 5.50 g. if the mixture contains 2.45 g k, what is the mass in grams of potassium hydroxide in the mixture? give the numerical part of your answer only; don't include the units.

Answer 1

Answer:

1.72 grams is the mass of potassium hydroxide in the mixture.

Explanation:

Total mass of the mixture = M = 5.50 g

Le the mass of KBr and KOH be x and y.

x + y = 5.50 g..[1]

Moles of KBr = $\frac{x}{119 g/mol}$

1 mol of KBr has 1 mol of potassium . Then $\frac{x}{119 g/mol}$ moles of KBr will have:

$1\times \frac{x}{119 g/mol}=\frac{x}{119 g/mol}$

Mass of potassium in x amount of KBr :

= $39 g/mol\times \frac{x}{119 g/mol}=0.33x$...[2]

Moles of KOH= $\frac{x}{56g/mol}$

1 mol of KOH has 1 mol of potassium . Then $\frac{y}{56g/mol}$ moles of KOH will have:

$1\times \frac{y}{56g/mol}=\frac{x}{56g/mol}$

Mass of potassium in y amount of KOH :

= $39 g/mol\times \frac{y}{56 g/mol}=0.70 y$...[3]

Mass of potassium in the mixture = 2.45 g

0.33x+ 0.70y=2.45 g..[4] (from [2] and [3] )

On solving [1] and [4] we get:

x = 3.78 g, y = 1.72 g

1.72 grams is the mass of potassium hydroxide in the mixture.

Answer 2

Answer is: mass in grams of KOH is 1,68.
m(KBr + KOH) = 5,50 g.
m(K) = 2,45 g.
M(KBr) = 39 g/mol + 80 g/mol = 119 g/mol.
M(KOH) = 39 g/mol + 16 g/mol + 1 g/mol = 56 g/mol.
ω(K) = 2,45 g ÷ 5,50 g.
ω(K) = 0,445 = 44,5%.
n(K) = m(K) ÷ M(K).
n(K) = 2,45 g ÷ 39 g/mol.
n(K) = 0,062 mol.
m(Br + OH) = 5,50 g - 2,45 g = 3,05 g.
n(K) = n(KBr + KOH) = 0,062 mol.
M(KBr + KOH) = 5,50 g ÷ 0,062 mol = 88,7 g/mol.
Set two equations:
1) m(KBr) + m(KOH) = 5,50 g.
2) m(KBr)/M(KBr) + m(KOH)/M(KOH) = 0,062 mol.
m(KOH) = 5,50 g - M(KBr), put that in second equation.
m(KBr) = 3,82 g.
m(KOH) = 5,50 g - 3,82 g = 1,68 g.