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3 years ago

Please help me and explanation would be really awesome thank you!

Chemistry

1 answer:

valentina_108 [34]3 years ago

4 0

Answer:

So the answer would be 10 moles

Explanation:

1) Start with the molecular formula for water: $H_{2} O!$

2) If there are 10 moles of water use a mole ratio to calculate the moles of oxygen it would produce.

(This question is... interesting... since they chose an element that is diatomic in free state so It could TECHNICALLY be two answers, moles of O or moles of $O_{2}$ )

The mole ratio is 1 moles of $H_{2}O$ to 1 moles of O. This is because the coefficient for oxygen in water is simple 1, so the ratio is 1:1.

3) that means if 10 moles of water decompose, they decompose into 10 moles of $H_{2}$ and 10 moles of O.

Extra:

About what I was saying before about the question being slightly interesting:

10 moles of pure oxygen is produced but free state oxygen exists as $O_{2}$ so it could possibly be 10 OR 5! However, notice it says elements. This leads me to believe the answer is 10 (monatomic oxygen) instead of 5 (free state/diatomic oxygen).

I hope this helps!

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8. A sample of potassium chlorate (KCIO,) was heated in a test tube and decomposed 2KC?(s) 302 (g) + 2KCIO, (s) The oxygen was c

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Answer:

Partial pressure of $O_{2}$ in the gas was 733 torr and mass of $KClO_{3}$ in the sample was 2.12 g.

Explanation:

a) Total pressure of gas = (partial pressure of water vapour)+(partial pressure of $O_{2}$ )

Here partial pressure of water vapour is 21 torr and total pressure of gas is 754 torr.

So, partial pressure of $O_{2}$ = (total pressure of gas)-(partial pressure of water vapour) = (754 torr) - (21 torr) = 733 torr

b) Lets assume that $O_{2}$ behaves ideally. Hence-

PV=nRT

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin

here P = 733 torr = $(733\times 0.001316)atm$ = 0.9646 atm

V = 0.65 L, R = 0.082 L.atm/(mol.K), T=(273+22)K = 295 K

So, $n=\frac{PV}{RT}$

= $\frac{(0.9646 atm)\times (0.65 L)}{(0.082 L.atm/(mol.K))\times (295 K)}$

= 0.0259 moles

As 3 moles of $O_{2}$ are produced from 2 moles of $KClO_{3}$ therefore 0.0259 moles of $O_{2}$ are produced from $(\frac{2\times 0.0259}{3})$ moles or 0.0173 moles of $KClO_{3}$ .

Molar mass of $KClO_{3}$ = 122.55 g

So mass of $KClO_{3}$ in sample = $(0.0173\times 122.55)g$

= 2.12 g

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