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oksian1 [2.3K]
4 years ago
15

Trace amounts of sulfur (S) in coal are burned in the presence of diatomic oxygen (O2) to form sulfur dioxide (SO2). Determine t

he minimum mass of oxygen required in the reactants and the mass of sulfur dioxide in the products when 1 kg of sulfur is burned.
Chemistry
1 answer:
Mashcka [7]4 years ago
3 0

Answer:

0.99 kg O₂

1.9 kg SO₂

Explanation:

Let's consider the reaction between sulfur and oxygen to form sulfur dioxide.

S + O₂ → SO₂

The mass ratio of S to O₂ is 32.07:32.00. The mass of oxygen required to react with 1 kg of sulfur is:

1 kg S × (32.00 kg O₂/32.07 kg S) = 0.998 kg O₂

The mass ratio of S to SO₂ is 32.07:64.07. The mass of sulfur dioxide formed when 1 kg of sulfur is burned is:

1 kg S × (64.07 kg SO₂/32.07 kg S) = 1.99 kg SO₂

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Answer:

O volume of the solution

Explanation:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / volume of solution  in L

For example:

if we dissolve the 1 mole of NaCl to make the solution of volume 2 L. The molarity of solution is,

M = 1 mol / 2 L

M = 0.5 M

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3 years ago
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6 0
3 years ago
Consider the reaction between iodine gas and chlorine gas to form iodine monochloride. A reaction mixture at 298.15k initially c
uranmaximum [27]

Step 1 : Write balanced chemical equation

The balanced chemical equation for the reaction between iodine gas and chlorine gas is given below.

I_{2}  (g) + Cl_{2}  (g) ----->  2 ICl (g)

Step 2 : Set up ICE table

We will set up an ICE table for the above reaction

Following points are considered while drawing ICE table

- The initial concentration of product is assumed as 0

- The change in concentration (C) is assumed as x. Change (x) is negative for reactants and positive for products

- The coefficients in balanced equation are considered while writing C values

Check attached file for ICE table

Step 3 : Set up equilibrium constant equation

The equation for equilibrium constant can be written as

K_{eq} = \frac{[ICl]^{2}}{[I_{2}][Cl_{2}]}

Step 4 : Solving for x

Keq at 298.15 K is given as 81.9

Let us plug in the equilibrium values (E) for I₂, Cl₂ and ICl from ICE table

81.9 = \frac{(2x)^{2}}{(0.437-x) (0.269-x)}

81.9 = \frac{(2x)^{2}}{x^{2} -0.706x + 0.118}

(2x)^{2} = 81.9 [ x^{2} -0.706x + 0.118]

4x^{2} = 81.9x^{2} -57.8x +9.66

77.9x^{2} -57.8x+9.66 = 0

Solving the above equation using quadratic formula we get

x = 0.488 or x = 0.254

The value 0.488 cannot be used because the change (C) cannot be greater that initial concentration of the reactants.

Therefore the change in concentration of the gases during the reaction is 0.254 M

Hence, x = 0.254 M

From the ICE table, we know that the equilibrium concentration of ICl is 2x

[ICl]_{eq} = 2 ( 0.254) = 0.508 M

The concentration of ICl when the reaction reaches equilibrium is 0.508 M

6 0
3 years ago
Predict how many grams of KCI is produced from 40 grams of K?
Nutka1998 [239]

Explanation:

firstly find for the molar mass of kcl and molar mass of k

and then

molar mass of k = x

molar mass of kcl= 40

cross mutiply and then simplify you will get your answer

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4 years ago
How does the ozone layer affect life on Earth?
zysi [14]
<span>d. filters out harmful ultraviolet radiation</span>
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