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Blizzard [7]
3 years ago
13

If 0.255 moles of AgNO₃ react with 0.155 moles of H₂SO₄ according to this UNBALANCED equation below, how many grams of Ag₂SO₄ co

uld be formed? AgNO₃(aq) + H₂SO₄ (aq) → Ag₂SO₄ (s) + HNO₃ (aq)
Chemistry
1 answer:
Ber [7]3 years ago
4 0

Answer: 6.162g of Ag2SO4 could be formed

Explanation:

Given;

0.255 moles of AgNO3

0.155 moles of H2SO4

Balanced equation will be given as;

2AgNO3(aq) + H2SO4(aq) -> Ag2SO4(s) + 2HNO3(aq)

Seeing that 2 moles of AgNO3 is required to react with 1 moles of H2SO4 to produce 1 mole of Ag2SO4,

Therefore the number of moles of Ag2SO4 produced is given by,

n(Ag2SO4) = 0.255 mol of AgNO3 ×

[0.155mol H2SO4 ÷ 2 mol AgNO3] x

[ 1 mol Ag2SO4 ÷ 1 mol H2SO4]

= 0.0198 mol of Ag2SO4.

mass = no of moles x molar mass

From literature, molar mass of Ag2SO4 = 311.799g/mol.

Thus,

Mass = 0.0198 x 311.799

= 6.162g

Therefore, 6.162g of Ag2SO4 could be formed

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A solution contains 50.0g of heptane (C7H16)and 50.0g of octane (C8H18) at 25 degrees C.The vapor pressures of pure heptane and
AleksandrR [38]

Answer:

a)Pheptane = 24.3 torr          

Poctane = 5.12 torr    

b)Ptotal vapor = 29.42 torr

c)  81 % heptane

    19 % octane

d) See explanation below

Explanation:

The partial pressure is given by Raoult´s law as:

Pa = Xa Pºa where Pa = partial pressure of component A

                               Xa = mole fraction of A

                               Pºa = vapor pressure of pure A

For a binary solution what we have to do is compute the partial  vapor pressure of each component and then add them together to get total vapor pressure.

In order to calculate the composition of the vapor  in part b), we will first calculate the mole fraction of each component in the vapor which is given by the relationship:

          Xa = Pa/Pt where Xa = mol fraction of  in the vapor

                                       Pa = partial pressure of A as calculated above

                                        Pt = total vapor pressure

Once we have mole fractions we can calculate the masses of the components for part c)    

a)                  

 MW heptane = 100.21 g/mol

 MW octane = 114.23 g/mol

mol heptane = 50.0 g / 100.21 g/mol = 0.50 mol

mol octane = 50.0 g/ 114.23 g/mol = 0.44 mol

mol total = 0.94 ⇒ Xa= 0.50/0.94 = 0.53 and

                             Xb= 0.44/0.94 = 0.47

Pheptane = 0.53 x 45.8 torr = 24.3 torr

Poctane = 0.47 x 10.9 torr = 5.12 torr

b) Ptotal = 24.3 torr +5.12 torr = 29.42 torr

c) We will call Y the mole fraction in the vapor to differentiate it from the mole fraction in solution

Y heptane (in the vapor) = 24.3 torr/ 29.42 torr = 0.83

Y octane (in the vapor) = 5.12 torr/ 29.42 torr = 0.17

d) To solve this part   we will assume that since the molecular weights are similar then having a mole fraction for heptane of 0.82, we could say that for every mole of mixture we have 0.82 mol heptane and 0.17 mol octane  and then we can calculate the masses:

0.82 mol x 100.21  g/mol = 82.2 g

0.17 mol x 114.23 g/mol =  19.4 g

total mass = 101.6

% heptane = 82.2 g/101.6g x 100 = 81 %

% octane = 19 %

There is another way to do this more exactly by calculating the average molecular weight of the mixture:

average MW = 0.83 (100.21 g/mol)  + 0.17 ( 114.23 g/mol ) = 102. 6 g/mol

and then  having a mol fraction of 0.83  means in 1 mol of mixture we have 0.83 mol heptane and 0.17 mol octane then the masses are:

mass heptane = 0.83 x 100.21 g/mol = 83.2 g

mass octane = 0.17 x  114.23 g/mol = 19.4 g

mass of mixture = 1 mol x MW mixture = 1 mol x 102.6 g/mol 102.6 g

% heptane = (83.2 g/ 102.6 g ) x 100 g = 81 %

% octane = 100 - 81 = 19 %

d)The composition of the vapor is different from the composition of the solution because the vapor is going to be richer in the more volatile compound in the solution which in this case is heptane ( 45.8  vs 10.9 torr).

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