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3 years ago

2KClO3 mc012-1.jpg 2KCl + 3O2 How many moles of oxygen are produced when 2 mol of potassium chlorate (KClO3) decompose?

Chemistry

2 answers:

Marizza181 [45]3 years ago

7 0

2KClO₃ = 2KCl + 3O₂

n(KClO₃)=2 mol

n(O₂)=3/2 *n(KClO₃)

n(O₂)=1.5 *2 mol = 3 mol

3 moles of oxygen are produced

Soloha48 [4]3 years ago

3 0

Answer: The moles of oxygen gas produced will be 3 moles.

Explanation:

For the given chemical reaction:

$2KClO_3\rightarrow 2KCl+3O_2$

By Stoichiometry of the reaction:

If 2 moles of potassium chlorate produces 3 moles of oxygen gas.

So, 2 moles of potassium chlorate will produce = $\frac{3}{2}]times 3=3$ moles of oxygen gas.

Hence, the moles of oxygen gas produced will be 3 moles.

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First, the question is incomplete, you are not giving the values of the temperatures and the pressure. However, I managed to find one similar question, and the given data is the temperature is lowered from 21 °C to -8°C, and the pressure decreased by 5%. If your data is different, you should only replace your data in the procedure, and you'll get an accurate result.

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Equalling both expressions and solving for V₂:

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V₂ = 264V₁ / 294*0.95

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%V = (V₁ - V₂ / V₁) * 100

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$q=mC\times \Delta T$ ......(1)

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As, the same amount of heat is getting absorbed in all the cases. So, the change in temperature will depend on the product of mass and specific heat.

For the given options:

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We are given:

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Putting values in equation 1, we get:

$200.0J=50.0g\times 0.449J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 0.449}=8.99^oC$

Change in temperature = 8.99°C

Option b: 50.0 g water, $C_{water}=4.18J/g^oC$

We are given:

$m=50.0g\\C_{water}=4.18J/g^oC$

Putting values in equation 1, we get:

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Option b: 25.0 g Pb, $C_{Pb}=0.128J/g^oC$

We are given:

$m=50.0g\\C_{Pb}=0.128J/g^oC$

Putting values in equation 1, we get:

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We are given:

$m=25.0g\\C_{Ag}=0.235J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.235J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.235}=34.04^oC$

Change in temperature = 34.04°C

Option e: 25.0 g granite, $C_{granite}=0.79J/g^oC$

We are given:

$m=25.0g\\C_{Fe}=0.79J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.79J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.79}=10.13^oC$

Change in temperature = 10.13°C

Hence, the smallest temperature change is shown by water.

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3 years ago