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DENIUS [597]
4 years ago
5

Decomposition Reactions (pls help!)

Chemistry
1 answer:
melamori03 [73]4 years ago
8 0

15. 1,1,1

16. 1,1,1

17. 1,1,1

18. 1,1,1

19. 1,1,1

20. 1,1,3

21. 2,2,3

22. 2,2,3

23. 1,1,1

24. 1,1,1

25. 2,4,3

26. 2,4,1

You should really learn to do these! They're actually very simple

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2 years ago
Calculate the Kc for the following reaction if an initial reaction mixture of 0.500 mole of CO and 1.500 mole of H2 in a 5.00 li
Lera25 [3.4K]

Answer:

4.41

Explanation:

Step 1: Write the balanced equation

CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)

Step 2: Calculate the respective concentrations

[CO]_i = \frac{0.500mol}{5.00L} = 0.100M

[H_2]_i = \frac{1.500mol}{5.00L} = 0.300M

[H_2O]_{eq} = \frac{0.198mol}{5.00L} = 0.0396M

Step 3: Make an ICE chart

        CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)

I       0.100      0.300        0            0

C         -x           -3x          +x          +x

E    0.100-x    0.300-3x     x            x

Step 4: Find the value of x

Since the concentration at equilibrium of water is 0.0396 M, x = 0.0396

Step 5: Find the concentrations at equilibrium

[CO] = 0.100-x = 0.100-0.0396 = 0.060 M

[H₂] = 0.300-3x = 0.300-3(0.0396) = 0.181 M

[CH₄] = x = 0.0396 M

[H₂O] = x = 0.0396 M

Step 6: Calculate the equilibrium constant (Kc)

Kc = \frac{[CH_4] \times [H_2O] }{[CO] \times [H_2]^{3} } = \frac{0.0396 \times 0.0396 }{0.060 \times 0.181^{3} } = 4.41

3 0
4 years ago
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