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Asked Oct 20, 2019
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If you place 7.05 g of aluminum in a beaker with 185 mL of 1.35 M KOH, what mass of KAl(OH)4 is produced?

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Expert Answer

Step 1

If 7.05 g of aluminum in a beaker is treated with 185 mL of 1.35 M KOH, the mass of KAl(OH)4 that is produced is to be calculated.

Step 2

Write the chemical equation for the reaction between aluminium and KOH-

Aluminium reacts with KOH in presence of water to produce KAl(OH)4 and hydrogen gas.

2A1+ 2KОH + 6H,0 > 2K[AI(ОН),]+ ЗН,
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2A1+ 2KОH + 6H,0 > 2K[AI(ОН),]+ ЗН,

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Step 3

Calculate the moles of aluminium-

Mass of Aluminium = 7.05 g

Mola...

Given mass of A
Moles of Al
Molar mass of Al
7.05 g
26.98 g/mol
. Moles of Al = 0.260
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Given mass of A Moles of Al Molar mass of Al 7.05 g 26.98 g/mol . Moles of Al = 0.260

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