Determine the number of moles in 258.45 grams of iron (III) chloride (FeCl3). Show and explain all
1 answer:
Answer:
Explanation:
To convert from grams to moles, we must use the molar mass. This can be found on the Periodic Table. First, find the molar mass of iron and chlorine.
- Fe: 55.84 g/mol
- Cl: 35.45 g/mol
Check the formula. There is a subscript of 3 after Cl, so there are 3 atoms of chlorine in 1 molecule. Multiply iron's molar mass by 3, then add iron's molar mass.
- FeCl₃: 55.84 + 3(35.45) = 55.84+106.35=162.19 g/mol
Use this number as a ratio.
Multiply by the given number of grams.
Flip the ratio so the grams of iron (III) chloride cancel.
The original measurement of grams has 5 significant figures, so our answer must have the same. For the number we calculated, that is the ten thousandth place.
258.45 grams is approximately 1.5935 moles of iron (III) chloride.
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Answer:
Aluminum metal
Explanation:
In order to properly answer this or a similar question, we need to know some basic rules about galvanic cells and standard reduction potentials.
First of all, your strategy would be to find a trusted source or the table of standard reduction potentials. You would then need to find the half-equations for aluminum and gold reduction:
Since we have a galvanic cell, the overall reaction is spontaneous. A spontaneous reaction indicates that the overall cell potential should be positive.
Since one half-equation should be an oxidation reaction (oxidation is loss of electrons) and one should be a reduction reaction (reduction is gain of electrons), one of these should be reversed.
Thinking simply, if the overall cell potential would be obtained by adding the two potentials, in order to acquite a positive number in the sum of potentials, we may only reverse the half-equation of aluminum (this would change the sign of E to positive):
Notice that the overall cell potential upon summing is:
Meaning we obey the law of galvanic cells.
Since oxidation is loss of electrons, notice that the loss of electrons takes place in the half-equation of aluminum: solid aluminum electrode loses 3 electrons to become aluminum cation.