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Diano4ka-milaya [45]
2 years ago
9

A mixture of 2.0 mol of CO(g)? and 2 mol of H2O?(g)? was allowed to come to equilibrium in a 1.0L flask at a high temperature. I

f Kc?=4.0, what is the molar concentration of H2(g) in the equilibrium mixture: CO(g) + H2O(g)\rightleftharpoons CO2(g) +H2(g)
Chemistry
1 answer:
Bess [88]2 years ago
7 0

For a mixture of 2.0 mol of CO(g) and 2 mol of H2O(g) was allowed to come to equilibrium in a 1.0L flask at a high temperature. the molar concentration of H2(g)  is mathematically given as

[H2] = x = 1.33 M

<h3>What is the molar concentration of H2(g) in the equilibrium mixture?</h3>

Generally, the equation for the Chemical reaction  is mathematically given as

Kc = [CO2]*[H2]/[CO]*[H2O]

Therefore

4.0 = (1*x)^2/(2-1*x)^2

x = 1.33

In conclusion, the molar concentration

[H2] = x = 1.33 M

Read more about  Chemical reaction

brainly.com/question/16416932

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Copper is formed when aluminum reacts with copper (II) sulfate in a single-replacement reaction. How many moles of copper can be
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Answer:

The answer to your question is the limiting reactant is CuSO₄ and 0.975 moles of Cu were obtained

Explanation:

moles of Copper = ?

mass of Aluminum = 29 g

mass of CuSO₄ = 156 g

Limiting reactant = ?

Balanced Chemical reaction

                  3 CuSO₄   +  2 Al   ⇒   3 Cu   +  Al₂(SO₄)₃

Calculate the moles of reactants

CuSO₄ = 64 + 32 + (16 x 4) = 160g

Al = 27 g

                160 g of CuSO₄  ----------------- 1 mol

                156 g                   -----------------  x

                      x = (156 x 1) / 160

                      x = 0.975 moles

               27 g of Al -------------------------- 1 mol

               29 g of Al -------------------------- x

                x = (29 x 1)/27

                x = 1.07 moles

Calculate proportions to find the limiting reactant

Theoretical     3 moles CuSO₄/2 moles Al = 1.5 moles

Experimental  0.975 moles CuSO₄/1.07 moles = 0.91

The experimental proportion was lower than the theoretical proportion that means that the limiting reactant is CuSO₄.      

                    3 moles of CuSO₄ ------------------ 3 moles of Cu

                 0.975 moles of CuSO₄ ---------------  x

                         x = (0.975 x 3)/3

                        x = 0.975 moles of Cu were obtained.        

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