Two samples of potassium iodide are decomposed into their constituent elements. The first sample produced 13.0 g of potassium an
1 answer:
Answer:
The mass of iodine produced will be 79.46 kg .
Explanation:
Let us calculate the mole ratio of potassium and iodine formed when first sample was decomposed.
Atomic mass of potassium =39 g/mol
Atomic mass of iodine = 127 g/mol
Moles of K produced =
Moles of I produced =
The mole ration of K :I = 1:1
The moles of K produced by second sample =
The same moles of I must have been produced.
The mass of Iodine produced = moles X atomic mass = 0.626 X 127=79.46 kg
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This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:
Substituting in our values, we get
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.