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Solved: What is the minimum volume of 3.0 M HBr required to completely dissolve 15.0 g
Chapter 14, Problem 92P(choose chapter or problem)
Questions & Answers
QUESTION:
Problem 92P
What is the minimum volume of 3.0 M HBr required to completely dissolve 15.0 g of potassium metal?
ANSWER:
Solution 92P :
Step 1:
Given :
Molarity of HBr (Hydrobromic acid) = 3.0 M
Amount of K(Potassium) metal = 15.0 g
Volume of HBr to completely dissolve Mg = ?
Let’s write the chemical equation for the given scenario :
2 K(s) + 2HBr → 2 KBr + H2(g)
From the above reaction, we see the ratio of reactants is 2 : 2.
Now, let’s find the number of moles of K and HBr :
Number of moles(n) =
Molar mass of K = 39.0983 g/mol.
Therefore, n =
= 0.3836 moles.
We are not given the mass of HBr, but from the reaction, we know that the ratio of K and HBr is 2 : 2. Hence, the moles of HBr will be 0.3836 moles.
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