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3 years ago

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Calculate the mass of carbon dioxide released when 185.000 g of copper carbonate [molar mass = 123.5 g/mol] is heated (assuming

that the reaction goes to completion). The molar mass of CO2 is 44.0 g/mol. Give your answer to three significant figures.

1 answer:

Vinvika [58]3 years ago

8 0

Answer:

The answer to your question is: 65.9 g released of CO2

Explanation:

MW CO2 = 44 g

MW CuCO3 = 123.5 g

CO2 released = ?

CuCO3 = 185 g

CuCO3 ⇒ CO2 + CuO

123.5 ----------- 44g

185 g ----------- x

x = (185 x 44) / 123.5

x = 65.9 g released of CO2

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<u><em>The frequency of the photon is 7.41*10¹⁶ Hz</em></u>

<u><em></em></u>

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Phosgene, COCl2, gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide w

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The question is incomplete, here is the complete question:

Phosgene, $COCl_2$ , gained notoriety as a chemical weapon in World War I. Phosgene is produced by the reaction of carbon monoxide with chlorine:

$CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)$

The value of $K_c$ for this reaction is 5.79 at 570 K. What are the equilibrium partial pressures of the three gases if a reaction vessel initially contains a mixture of the reactants in which $p_{CO}=p_{Cl_2}=0.265atm$ and $p_{COCl_2}=0.000atm$ ?

<u>Answer:</u> The equilibrium partial pressure of CO, $Cl_2\text{ and }COCl_2$ is 0.257 atm, 0.257 atm and 0.008 atm respectively.

<u>Explanation:</u>

The relation of $K_c\text{ and }K_p$ is given by:

$K_p=K_c(RT)^{\Delta n_g}$

$K_p$ = Equilibrium constant in terms of partial pressure

$K_c$ = Equilibrium constant in terms of concentration = 5.79

$\Delta n_g$ = Difference between gaseous moles on product side and reactant side = $n_{g,p}-n_{g,r}=1-2=-1$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

T = Temperature = 570 K

Putting values in above equation, we get:

$K_p=5.79\times (0.0821\times 570)^{-1}\\\\K_p=0.124$

We are given:

Initial partial pressure of CO = 0.265 atm

Initial partial pressure of chlorine gas = 0.265 atm

Initial partial pressure of phosgene = 0.00 atm

The given chemical equation follows:

$CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)$

<u>Initial:</u> 0.265 0.265

<u>At eqllm:</u> 0.265-x 0.265-x x

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{COCl_2}}{p_{CO}\times p_{Cl_2}}$

Putting values in above equation, we get:

$0.124=\frac{x}{(0.265-x)\times (0.265-x)}\\\\x=0.0082,8.59$

Neglecting the value of x = 8.59 because equilibrium partial pressure cannot be greater than initial pressure

So, the equilibrium partial pressure of CO = $(0.265-x)=(0.265-0.008)=0.257atm$

The equilibrium partial pressure of $Cl_2=(0.265-x)=(0.265-0.008)=0.257atm$

The equilibrium partial pressure of $COCl_2=x=0.008atm$

Hence, the equilibrium partial pressure of CO, $Cl_2\text{ and }COCl_2$ is 0.257 atm, 0.257 atm and 0.008 atm respectively.

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