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givi [52]
3 years ago
7

How man grams of cl2 are consumed to produce 12.0 g of KCl

Chemistry
1 answer:
Korvikt [17]3 years ago
8 0

Answer:

5.71 g

Explanation:

Step 1: Write the balanced equation

2 K + Cl₂ ⇒ 2 KCl

Step 2: Calculate the moles corresponding to 12.0 g of KCl

The molar mass of KCl is 74.55 g/mol.

12.0 g × 1 mol/74.55 g = 0.161 mol

Step 3: Calculate the moles of Cl₂ needed to produce 0.161 moles of KCl

The molar ratio of Cl₂ to KCl is 1:2. The moles of Cl₂ needed are 1/2 × 0.161 mol = 0.0805 mol

Step 4: Calculate the mass corresponding to 0.0805 moles of Cl₂

The molar mass of Cl₂ is 70.91 g/mol.

0.0805 mol × 70.91 g/mol = 5.71 g

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The equilibrium constant in terms of pressure. Kp for the following process is 0.179 at 50 °C. It increases to 0.669 at 86 °C. C
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a) ΔHvap=35.3395 kJ/mol

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Explanation:

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When

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The Clasius-Clapeyron equation is:

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

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ln(\frac{1atm}{0.179atm}) =-\frac{35339.5 J/mol}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{T_2}-\frac{1}{323.15K})

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