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tatiyna
3 years ago
11

when 45 grams of copper (ii) carbonate are decomposed with heat how many grams of carbon dioxide will be produced

Chemistry
1 answer:
Maksim231197 [3]3 years ago
8 0

Answer:

16.02 g

Explanation:

the balanced equation for the decomposition of CuCO₃ is as follows

CuCO₃ --> CuO + CO₂

molar ratio of CuCO₃ to CO₂ is 1:1

number of CuCO₃ moles decomposed - 45 g / 123.5 g/mol = 0.364 mol

according to the molar ratio

1 mol of CuCO₃ decomposes to form 1 mol of CO₂

therefore 0.364 mol of  CuCO₃ decomposes to form 0.364 mol of CO₂

number of CO₂ moles produced - 0.364 mol

therefore mass of CO₂ produced - 0.364 mol x 44 g/mol = 16.02 g

16.02 g of CO₂ produced

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There was a part missing. I think this is the whole question:

<em>Before any reaction occurs, the concentration of A in the reaction below is 0.0510 M. What is the equilibrium constant if the concentration of A at equilibrium is 0.0153 M?</em>

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<em>Remember to use correct significant figures in your answer. Do not include units in your response.</em>

First, we have to make an ICE Chart, which stands for initial, change and equilibrium. We will call "x" unknown concentrations.

        A (aq) ⇌ 2B (aq) + C (aq)

I       0.0510       0            0

C         -x          +2x          +x

E    0.0510-x     2x            x

Since the concentration at equilibrium of A is 0.0153 M, we get

0.0510 - x = 0.0153 \\x = 0.0357

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[A]e = 0.0153 M \\[B]e = 2x = 2(0.0357) = 0.0714 M \\[C]e = x = 0.0357 M \\

The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients.

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The equilibrium constant for this reaction at equilibrium is 0.0119.

You can learn more about equilibrium here: brainly.com/question/4289021

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