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Serggg [28]
3 years ago
5

In the reaction MgCl2 + 2KOH → Mg(OH)2 + 2KCI, how many grams of KOH

Chemistry
1 answer:
dsp733 years ago
4 0

Answer:

The correct option is;

D. (2)(56 g)

Explanation:

MgCl₂ + 2KOH → Mg(OH)₂ + 2KCl

From the balanced chemical reaction equation, we have;

One mole of MgCl₂ reacts with  two moles of KOH to produce one mole of Mg(OH)₂ and 2 moles of KCl

Therefore, the number of moles of KOH that react with one mole of KCl = 2 moles

The mass, m, of the two moles of KOH = Number of moles of KOH × Molar mass of KOH

The molar mass of KOH = 56.1056 g/mol

∴ The mass, m, of the two moles of KOH = 2 moles × 56.1056 g/mol = 112.2112 grams

The amount in grams of KOH that react with one mole of MgCl₂ = 112.2112 grams ≈ 112 grams =  (2)(56 g).

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balu736 [363]
The formula is  m = D x V
D = <span>13.69 g/cm^3.
</span>V = <span>15.0 cm^3 
the mass of the liquid mercury is m= </span>13.69 g/cm^3 x 15.0 cm^3 = 195g
the molar mass of Hg is 200,
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but we know that
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0.97 mole=?

6.022 E23 atoms x 0.97 / 1 mole = 5.84 E23 atoms
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1.81 x 10²⁴ atoms

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