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cupoosta [38]
3 years ago
7

Use the equation: 2 kclo3 2 kcl + 3 o2 to write the mole ratio needed to calculate: (a) moles of kcl produced from 3.5 moles of

kclo3 mol kcl mol kclo3 (b) moles of kcl produced when 4.5 moles of o2 are produced. mol kcl mol o2
Chemistry
1 answer:
poizon [28]3 years ago
3 0

Answer:-

(a) 3.5

(b) 3

Explanation:-

2KClO3 --> 2KCl + 3 O2

From the equation we see that 2 moles of KClO3 gives 2 moles of KCl.

So 3.5 moles of KClO3 will give 3 moles of KCl.

Again

3 moles of O2 are produced with 2 moles of KCl.

If 4.5 moles of O2 produced then

moles of KCl = 4.5 x 2/3

=3

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8 0
2 years ago
The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p
saul85 [17]

Answer : The vapor pressure of bromine at 10.0^oC is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of bromine at 10.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 10.0^oC=273+10.0=283.0K

T_2 = normal boiling point of bromine = 58.8^oC=273+58.8=331.8K

\Delta H_{vap} = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})

P_1=0.1448atm

Hence, the vapor pressure of bromine at 10.0^oC is 0.1448 atm.

4 0
3 years ago
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