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3 years ago

Mg(OH)2 + 2 HBr à MgBr2 + 2 H2O

Chemistry

1 answer:

AnnyKZ [126]3 years ago

5 0

Explanation:

The balanced equation of the reaction is given as;

Mg(OH)2 (s) + 2 HBr (aq) → MgBr2 (aq) + 2 H2O (l)

1. How many grams of MgBr2 will be produced from 18.3 grams of HBr?

From the reaction;

2 mol of HBr produces 1 mol of MgBr2

Converting to masses using;

Mass = Number of moles * Molar mass

Molar mass of HBr = 80.91 g/mol

Molar mass of MgBr2 = 184.113 g/mol

This means;

(2 * 80.91 = 161.82g) of HBr produces (1 * 184.113 = 184.113g) MgBr2

18.3g would produce x

161.82 = 184.113

18.3 = x

x = (184.113 * 18.3 ) / 161.82 = 20.8 g

2. How many moles of H2O will be produced from 18.3 grams of HBr?

Converting the mass to mol;

Number of moles = Mass / Molar mass = 18.3 / 80.91 = 0.226 mol

From the reaction;

2 mol of HBr produces 2 mol of H2O

0.226 mol would produce x

2 =2

0.226 = x

x = 0.226 * 2 / 2 = 0.226 mol

3. How many grams of Mg(OH)2 are needed to completely react with 18.3 grams of HBr?

From the reaction;

2 mol of HBr reacts with 1 mol of Mg(OH)2

18.3g of HBr = 0.226 mol

2 = 1

0.226 = x

x = 0.226 * 1 /2

x = 0.113 mol

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The question is incomplete, here is the complete question:

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Explanation:

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R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

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Putting values in above equation, we get:

$0.901atm\times 50.0L=n\times 0.0821\text{ L. atm}mol^{-1}K^{-1}\times 923K\\\\n=\frac{0.901\times 50.0}{0.0821\times 923}=0.594mol$

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