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lisabon 2012 [21]
3 years ago
9

Calculate the percent composition of muscovite mica. Its chemical

Chemistry
1 answer:
Vikki [24]3 years ago
7 0

Answer:

Explanation:

(KF)2(Al2O3)3(SiO2)6(H2O)

molecular formula

K₂F₂Al₆Si₆O₂₂H₂

2 x 39 + 2 x 19 + 6 x 27 + 6 x 28 + 22 x 16 + 2 x 2

=  78 + 38 + 162+ 168+ 352+ 4

= 802

Percentage of K = 78 x 100 / 802

= 9.72 %

Percentage of F = 38 x 100 / 802

= 4.74 %

Percentage of Al = 162 x 100 / 802

= 20.2  %

Percentage of Si = 168 x 100 / 802

= 20.9  %

Percentage of O = 352 x 100 / 802

= 43.9 %

Percentage of H = 4 x 100 / 802

= .54  %

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Please help...awarding brainliest if correct and lots of points.
abruzzese [7]

Silver chloride produced : = 46.149 g

Limiting reagent : CuCl2

Excess remains := 3.74 g

<h3>Further explanation</h3>

Reaction

silver nitrate + copper(II) chloride ⇒ silver chloride + copper(II) nitrate

Required

silver chloride produced

limiting reagent

excess remains

Solution

Balanced equation

2AgNO3 (aq) + CuCl2 (s) → 2AgCl(s) + Cu(NO3)2(aq)

mol AgNO3 :

= 58.5 : 169,87 g/mol

= 0.344

mol CuCl2 :

=21.7 : 134,45 g/mol

= 0.161

mol ratio : coefficient of AgNO3 : CuCl2 :

= 0.344/2 : 0.161/1

= 0.172 : 0.161

CuCl2  as a limiting reagent

mol AgCl :

= 2/1 x 0.161

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