What is the minimum volume of 2.6 M HBr required to completely dissolve 35.0 g of potassium metal?
.......2HBr + 2K ==> 2KBr + H2
mols K = grams/molar mass
Using the coefficients in the balanced equation, convert mols K to mols HBr.
Now convert mols HBr to volume using
M HBr = mols HBr/L HBr. You know M and mols, solve for L.
Well, if we're talking about dissolving potassium metal, we might need a bit more than just HBr. We might also need the Magic Potion of Extraordinary Dissolution! Just kidding. But seriously, this sounds like a chemistry question, and I'm here to clown around. So let's get started!
To solve this problem, we need to convert the grams of potassium metal into moles by using its molar mass. Then we can use the balanced chemical equation to determine the stoichiometry between HBr and potassium. Finally, we can use the molarity of HBr to calculate the volume needed.
Are you ready to embark on this scientific journey with me, or should I just bring out some balloons and juggling balls?
To determine the minimum volume of 2.6 M HBr required to completely dissolve 35.0 g of potassium metal, we need to calculate the moles of potassium (K) and then use the stoichiometry of the balanced chemical equation.
The balanced chemical equation for the reaction between potassium and hydrobromic acid (HBr) is:
2 K + 2 HBr ⟶ 2 KBr + H2
From the equation, we can see that 2 moles of potassium metal react with 2 moles of hydrobromic acid to produce 2 moles of potassium bromide and 1 mole of hydrogen gas.
Step 1: Calculate the moles of potassium (K) in 35.0 g.
To do this, we need to know the molar mass of potassium (K), which is 39.10 g/mol.
moles of K = mass of K / molar mass of K
moles of K = 35.0 g / 39.10 g/mol
moles of K = 0.895 mol (rounded to three decimal places)
Step 2: Determine the moles of hydrobromic acid (HBr) required.
Since the stoichiometric ratio between K and HBr is 2:2, the moles of HBr required will be the same as the moles of K.
moles of HBr required = moles of K = 0.895 mol
Step 3: Calculate the volume of 2.6 M HBr required.
The concentration of HBr is given as 2.6 M. This means that there are 2.6 moles of HBr in a liter of the solution (1 M = 1 mol/L).
volume of HBr required = moles of HBr required / concentration of HBr
volume of HBr required = 0.895 mol / 2.6 mol/L
volume of HBr required ≈ 0.344 L or 344 mL (rounded to three significant figures)
Therefore, the minimum volume of 2.6 M HBr required to completely dissolve 35.0 g of potassium metal is approximately 0.344 L or 344 mL.
To determine the minimum volume of 2.6 M HBr (hydrobromic acid) required to completely dissolve 35.0 g of potassium metal, you can follow these steps:
Step 1: Write a balanced chemical equation
First, we need to write the balanced chemical equation for the reaction between HBr and potassium metal. The balanced equation is:
2 K + 2 HBr -> 2 KBr + H2
This equation shows that two moles of potassium react with two moles of hydrobromic acid to produce two moles of potassium bromide and one mole of hydrogen gas.
Step 2: Convert grams of potassium to moles
Use the molar mass of potassium (K) to convert the given mass of potassium (35.0 g) to moles. The molar mass of potassium is 39.10 g/mol.
35.0 g K * (1 mol K / 39.10 g K) = 0.895 mol K
Step 3: Determine the stoichiometry
Based on the balanced chemical equation, the stoichiometric ratio between K and HBr is 2:2. This means that for every 2 moles of potassium, we need 2 moles of HBr for a complete reaction.
Step 4: Calculate the volume of HBr
To find the volume of 2.6 M HBr required, we can use the equation:
moles = Molarity * Volume (in liters)
Since we want to find the volume, we can rearrange the equation as:
Volume (in liters) = moles / Molarity
Volume (in liters) = (0.895 mol K / 2) / (2.6 mol/L)
Volume (in liters) = 0.172 L
Step 5: Convert the volume to milliliters (mL)
To convert the volume from liters to milliliters, use the conversion factor: 1 L = 1000 mL.
Volume (in mL) = 0.172 L * 1000 mL/L = 172 mL
Therefore, the minimum volume of 2.6 M HBr required to completely dissolve 35.0 g of potassium metal is 172 mL.