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nignag [31]
2 years ago
14

How many grams of Fe can be produced when 5.50 g of Fe2O3 reacts?

Chemistry
1 answer:
vichka [17]2 years ago
7 0

Answer:

3,85 g of Fe

Explanation:

1- The first thing to do is calculate the molar mass of the Fe2O3 compound. With the help of a periodic table, the weights of the atoms are searched, and the sum is made:

Molar mass of Fe2O3 = (2 x mass of Fe) + (3 x mass of O) = 2 x 55.88 g + 3 x 15.99 g = 159.65 g / mol

Then, one mole of Fe2O3 has a mass of 159.65 grams.

2- Then, the relationship between the Fe2O3 that will react and the iron to be produced. With the previous calculation, we can say that with one mole of Fe2O3, two moles of Fe can be produced. Passing this relationship to the molar masses, it would be as follows:

1 mole of Fe2O3_____ 2 moles of Fe

159.65 g of Fe2O3_____ 111.76 g of Fe

3- Finally, the calculation of the mass that can be produced of Fe is made, starting from 5.50 g of Fe2O3

159.65 g of Fe2O3 _____ 111.76 g of Fe

5.50 g of Fe2O3 ______ X = 3.85 g of Fe

<em>Calculation: 5.50 g x 111.76 g / 159.65 g = 3.85 g </em>

The answer is that 3.85 g of Fe can be produced when 5.50 g of Fe2O3 react

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Sonja [21]

Answer : The initial concentration of HI and concentration of HI at equilibrium is, 0.27 M and 0.386 M  respectively.

Solution :  Given,

Initial concentration of H_2 and I_2 = 0.11 M

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The given equilibrium reaction is,

    H_2(g)+I_2(g)\rightleftharpoons 2HI(g)

Initially               0.11   0.11            C

At equilibrium  (0.11-x) (0.11-x)   (C+2x)

As we are given that:

Concentration of H_2 and I_2 at equilibrium = 0.052 M  = (0.11-x)

The expression of K_c will be,

K_c=\frac{[HI]^2}{[H_2][I_2]}

54.3=\frac{(C+2(0.058))^2}{(0.052)\times (0.052)}

By solving the terms, we get:

C = 0.27 M

Thus, initial concentration of HI = C = 0.27 M

Thus, the concentration of HI at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M

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