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Answer:
The correct answer is option d.
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of credit
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<u>Answer:</u>
<u>For 1:</u> The amount of potassium iodate that were titrated is moles
<u>For 2:</u> The amount of sodium thiosulfate required is moles
<u>Explanation:</u>
- <u>For 1:</u>
To calculate the number of moles for given molarity, we use the equation:
Molarity of solution = 0.0100 M
Volume of solution = 25 mL
Putting values in above equation, we get:
Hence, the amount of potassium iodate that were titrated is moles
- <u>For 2:</u>
The chemical equation for the reaction of potassium iodate and sodium thiosulfate follows:
By Stoichiometry of the reaction:
2 moles of potassium iodate reacts with 1 mole of sodium thiosulfate
So, 0.00025 moles of potassium iodate will react with = of sodium thiosulfate
Hence, the amount of sodium thiosulfate required is moles
N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g)
1) to calculate the limiting reactant you need to pass grams to moles.
<span> moles is calculated by dividing mass by molar mass
</span>
mass of N2O4: 50.0 g
molar mass of <span>N2O4 = 92.02 g/mol
</span><span>molar mass of N2H4 = 32.05 g/mol.
</span>mass of N2H4:45.0 g
moles N2O4=50.0/92.02 g/mol= 0,54 mol of N2O4
moles N2H4= 45/32.05 g/mol= 1,40 mol of <span><span>N2H4
</span> 2)</span>
By looking at the balanced equation, you can see that 1 mol of N2O4 needs 2 moles of N2H4 to fully react . So to react 0,54 moles of N2O4, you need 2x0,54 moles of <span>N2H4 moles
</span><span>N2H4 needed = 1,08 moles.
You have more that 1,08 moles </span><span>N2H4, so this means the limiting reagent is not N2H4, it's </span>N2O4. The molecule that has molecules that are left is never the limiting reactant.
3) 1 mol of N2O4 reacting, will produce 3 mol of N2 (look at the equation)
There are 0,54 mol of N2O4 available to react, so how many moles will produce of N2?
1 mol N2O4------------3 mol of N2
0,54 mol N2O4--------x
x=1,62 mol of N2
4) the only thing left to do is convert the moles obtained, to grams.
We use the same formula as before, moles equal to mass divided by molar mass.
(molar mass of N2= 28)
1,62 mol of N2= mass/ 28
mass of N2= 45,36 grams
1) to calculate the limiting reactant you need to pass grams to moles.
<span> moles is calculated by dividing mass by molar mass
</span>
mass of N2O4: 50.0 g
molar mass of <span>N2O4 = 92.02 g/mol
</span><span>molar mass of N2H4 = 32.05 g/mol.
</span>mass of N2H4:45.0 g
moles N2O4=50.0/92.02 g/mol= 0,54 mol of N2O4
moles N2H4= 45/32.05 g/mol= 1,40 mol of <span><span>N2H4
</span> 2)</span>
By looking at the balanced equation, you can see that 1 mol of N2O4 needs 2 moles of N2H4 to fully react . So to react 0,54 moles of N2O4, you need 2x0,54 moles of <span>N2H4 moles
</span><span>N2H4 needed = 1,08 moles.
You have more that 1,08 moles </span><span>N2H4, so this means the limiting reagent is not N2H4, it's </span>N2O4. The molecule that has molecules that are left is never the limiting reactant.
3) 1 mol of N2O4 reacting, will produce 3 mol of N2 (look at the equation)
There are 0,54 mol of N2O4 available to react, so how many moles will produce of N2?
1 mol N2O4------------3 mol of N2
0,54 mol N2O4--------x
x=1,62 mol of N2
4) the only thing left to do is convert the moles obtained, to grams.
We use the same formula as before, moles equal to mass divided by molar mass.
1,62 mol of N2= mass/ 28
mass of N2= 45,36 grams