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Vesnalui [34]
3 years ago
11

How many moles are in 128.9 grams of Cr2(SO3)2?

Chemistry
1 answer:
Dominik [7]3 years ago
7 0

Answer: 0.5 moles

Explanation:

Cr2(SO3)2 is the chemical formula for chromium sulphate.

Given that,

Amount of moles of Cr2(SO3)2 (n) = ?

Mass of Cr2(SO3)2 in grams = 128.9g

For molar mass of Cr2(SO3)2, use the atomic masses:

Chromium, Cr = 52g;

Sulphur, S = 32g;

Oxygen, O = 16g

Cr2(SO3)2 =

(52g x 2) + [(32g + 16g x 3) x 2]

= 104g + [(32g + 48g) x 2]

= 104g + [80g x 2]

= 104g + 160g

= 264g/mol

Since, n = mass in grams / molar mass

n = 128.9g / 264g/mol

n = 0.488 mole [Round the value of n to the nearest tenth which is 0.5

Thus, there are 0.5 moles in 128.9 grams of Cr2(SO3)2

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If 3.0 atm of pure HN3(g) is decomposed initially, what is the final total pressure in the reaction container? What are the part
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This is an incomplete question, here is a complete question.

Hydrogen azide, HN₃, decomposes on heating by thefollowing unbalanced reaction:

HN_3(g)\rightarrow N_2(g)+H_2(g)

If 3.0 atm of pure HN₃ (g) is decomposed initially,what is the final total pressure in the reaction container? Whatare the partial pressures of nitrogen and hydrogen gas? Assume thatthe volume and temperature of the reaction container are constant.

Answer : The partial pressure of N_2 and H_2 gases are, 4.5 atm and 1.5 atm respectively.

Explanation :

The given unbalanced chemical reaction is:

HN_3(g)\rightarrow N_2(g)+H_2(g)

This reaction is an unbalanced chemical reaction because in this reaction number of hydrogen and nitrogen atoms are not balanced on both side of the reaction.

In order to balance the chemical equation, the coefficient '2' put before the HN_3 and the coefficient '3' put before the N_2 then we get the balanced chemical equation.

The balanced chemical reaction will be,

2HN_3(g)\rightarrow 3N_2(g)+H_2(g)

As we are given:

The pressure of pure HN_3 = 3.0 atm

p_{Total}=2\times p_{HN_3}=2\times 3.0atm=6.0atm

From the reaction we conclude that:

Number of moles of N_2 = 3 mol

Number of moles of H_2 = 1 mol

Now we have to calculate the mole fraction of N_2 and H_2

\text{Mole fraction of }N_2=\frac{\text{Moles of }N_2}{\text{Moles of }N_2+\text{Moles of }H_2}=\frac{3}{3+1}=0.75

and,

\text{Mole fraction of }H_2=\frac{\text{Moles of }H_2}{\text{Moles of }N_2+\text{Moles of }H_2}=\frac{1}{3+1}=0.25

Now we have to calculate the partial pressure of N_2 and H_2

According to the Raoult's law,

p_i=X_i\times p_T

where,

p_i = partial pressure of gas

p_T = total pressure of gas  = 6.0 atm

X_i = mole fraction of gas

p_{N_2}=X_{N_2}\times p_T

p_{N_2}=0.75\times 6.0atm=4.5atm

and,

p_{H_2}=X_{H_2}\times p_T

p_{H_2}=0.25\times 6.0atm=1.5atm

Thus, the partial pressure of N_2 and H_2 gases are, 4.5 atm and 1.5 atm respectively.

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