Using the half-reaction method, balance the following redox reaction occurring in acidic solution:
1 answer:
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The amount of KCl that would be formed will be 478.8 grams
<h3>Stoichiometric calculations</h3>From the equation of the reaction, the mole ratio of K to is 2:1.
Mole of 309 g potassium = 309/39 = 7.92 moles
Mole of 228 g Cl2 = 228/71 = 3.21 moles
Thus, potassium is in excess while Cl2 is limiting.
Mole ratio of Cl2 and KCl = 1:2
Equivalent mole of KCl = 3.21 x 2 = 6.42 moles
Mass of 6.42 moles KCl = 6.42 x 74.55 = 478.8 grams
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Answer:
The answer is
<h2>2.64 g/mL</h2>Explanation:
The density of a substance can be found by using the formula
From the question
mass = 611 g
volume = final volume of water - initial volume of water
volume = 256.7 - 25.1 = 231.6 mL
The density of the metal is
We have the final answer as
<h3>2.64 g/mL</h3>Hope this helps you