Answer:
Explanation:
Hello,
In this case, for the given reactants we identify the following chemical reaction:
Thus, we evidence a 1:1 molar ratio between KOH and HCl, therefore, for the complete neutralization we have equal number of moles, that in terms of molarities and volumes become:
Hence, we compute the volume of HCl as shown below:
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<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
Molarity of NaI solution = 0.130 M
Volume of solution = 0.400 L
Putting values in above equation, we get:
The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:
The precipitate (insoluble salt) formed is lead (II) iodide
By Stoichiometry of the reaction:
2 moles of NaI produces 1 mole of lead (II) iodide
So, 0.52 moles of NaI will produce = of lead (II) iodide
To calculate the number of moles, we use the equation:
Moles of lead (II) iodide = 0.26 moles
Molar mass of lead (II) iodide = 461.1 g/mol
Putting values in above equation, we get:
Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams
Answer: 8.691 mols of CO₂
Explanation:
To find the number of moles in a given grams, you want to use the molar mass.
Let's first find the molar mass of CO₂.
Carbon's molar mass is 12.011 g/mol
Oxygen's molar mass is 15.999 g/mol
To find molar mass of CO₂, we want to add up the molar mass of carbon and oxygen. Remember, there are 2 Oxygens so we need to mulitply that by 2.
12.011+2(15.999)=44.009 g/mol
Now that we have molar mass, we can convert 382.5 g to mols.
There are about 8.691 mols of CO₂.