A 25.00-mL aliquot of a nitric acid solution of unknown concentration is pipetted into a 125-mL Erlenmeyer flask and 2 drops of
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<u>Answer:</u> The theoretical yield of iron(III) sulfate is 26.6 grams
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of iron(III) phosphate = 20.00 g
Molar mass of iron(III) phosphate = 150.82 g/mol
Putting values in equation 1, we get:
The given chemical equation follows:
As, sodium sulfate is present in excess. So, it is considered as an excess reagent.
Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate
So, 0.133 moles of iron(III) phosphate will produce = of iron(III) sulfate
Now, calculating the mass of iron(III) sulfate from equation 1, we get:
Molar mass of iron(III) sulfate = 399.9 g/mol
Moles of iron(III) sulfate = 0.0665 moles
Putting values in equation 1, we get:
Hence, the theoretical yield of iron(III) sulfate is 26.6 grams
Answer: 116 g of copper
Explanation:
where Q= quantity of electricity in coloumbs
I = current in amperes = 24.5A
t= time in seconds = 4.00 hr = (1hr=3600s)
of electricity deposits 63.5 g of copper.
352800 C of electricity deposits = of copper.
Thus 116 g of Cu(s) is electroplated by running 24.5A of current
Thus remaining in solution = (0.1-0.003)=0.097moles