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Anastasy [175]
3 years ago
12

If a maximum of 5.60g of a gas at 2.50 atm of pressure dissolves in 3.5-L of water, what will the solubility be if the pressure

Chemistry
1 answer:
KonstantinChe [14]3 years ago
6 0
According to Henry's law C = k P
where C = concentration of gas (g/L)
k : Henry's constant
P : pressure of gas above the solution
concentration of gas = 5.6 g / 3.5 L = 1.6 g/L
From Henry's law:
C₁ / C₂ = P₁ / P₂
C₂ = C₁P₂ / P₁ = (1.6 * 5 atm) / 2.5 atm = 3.2 g/L
so the 3.5 L will contain 11.2 g gas

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Hey there!:

Molar mass of Mg(OH)2 = 58.33 g/mol


number of moles Mg(OH)2 :

moles of Mg(OH)2 = 30.6 / 58.33 =>  0.5246 moles

Molar mass of H3PO4 =  97.99 g/mol

number of moles H3PO4:

moles of Mg(OH)2 = 63.6 / 97.99 => 0.649 moles

Balanced chemical equation is:


3 Mg(OH)2 + 2 H3PO4 --->  Mg3(PO4)2 + 6 H2O


3 mol of Mg(OH)2 reacts with 2 mol of H3PO4  ,for 0.5246 moles of Mg(OH)2, 0.3498 moles of H3PO4 is required , but we have 0.649 moles of H3PO4, so, Mg(OH)2 is limiting reagent !

Now , we will use Mg(OH)2 in further calculation  .

Molar mass of Mg3(PO4)2 = 262.87 g/mol

According to balanced equation  :

mol of Mg3(PO4)2 formed = (1/3)* moles of Mg(OH)2


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use :

mass of Mg3(PO4)2 = number of mol * molar mass


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= 46 g of Mg3(PO4)2

Therefore:

% yield = actual mass * 100 / theoretical mass

% = 34.7 * 100 / 46

% = 3470 / 46

= 75.5%


Hope that helps!




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