Write the complete redox and the two half-reactions for the reaction of aluminum
ID: 949787 • Letter: W
Question
Write the complete redox and the two half-reactions for the reaction of aluminum and copper. Which species is oxidizing agent? The reducing agent? How many electrons were transferred?
A compound containing xenon and fluorine, Xex,Fy, was produced by shining bright sunlight on a gaseous mixture of 0.439 g of Xe and excess fluorine. If all the xenon reacts, and 0.693 grams of the compound are formed, what is the empirical formula of the compound?
A 0.158 gram sample of a salt, comprised of barium and one of the halide ions, was dissolved in water, and an excess of sulfuric acid was added to form barium sulfate, BaSO4. After filtering and drying, the solid BaSO4 weighed 0.124 grams. Determine the identity of the halide and the formula of the barium halide.
Explanation / Answer
Write the complete redox and the two half-reactions for the reaction of aluminum and copper. Which species is oxidizing agent? The reducing agent? How many electrons were transferred?
The complete redox reaction is between aluminum metal and copper ions:
2Al(s) + 3Cu2+ --> 2Al^3+ + 3Cu(s)
2(Al --> Al3+ + 3e-) -- Oxidation half-reaction
3(Cu2+ + 2e- --> Cu) -- Reduction half-reaction
---------------------------------
2Al(s) + 3Cu2+ --> 2Al^3+ + 3Cu(s)
In this reaction Al is oxidized because its oxidation number changed from 0 to +3 . It provides electron for the reduction means it is a reducing agent.
In this reaction Cu is reduced because its oxidation number changed from 2+ to 0 . It accepts electron for the oxidation means it is an oxidizing agent.
In this reaction total 6 e are transferred.
A compound containing xenon and fluorine, Xex,Fy, was produced by shining bright sunlight on a gaseous mixture of 0.439 g of Xe and excess fluorine. If all the xenon reacts, and 0.693 grams of the compound are formed, what is the empirical formula of the compound?
moles Xe = 0.439 g/ 131.29 g/mol= 0.0033
amount of F = 0.693 - 0.439 = 0.254 g
Moles F =0.254 g / 18.9984 g/mol=0.013
Now calculate the mole ratio in the compound:
0.0033/0.0033=1.0 means one Xe
0.013/ 0.0033 = 4.0 means 4 F
Thus the empirical formula is XeF4.
A 0.158 gram sample of a salt, comprised of barium and one of the halide ions, was dissolved in water, and an excess of sulfuric acid was added to form barium sulfate, BaSO4. After filtering and drying, the solid BaSO4 weighed 0.124 grams. Determine the identity of the halide and the formula of the barium halide.
The common reaction between BaX2 and H2SO4 is following:
BaX2 + H2SO4 ----->BaSO4 + 2HX
molar mass of H2SO4 = 98 g/mol
molar mass of BaSO4 = 233.3 g/mol
molar mass of all barium halides are following:
molar mass of BaF2= 175.3 g/mol
molar mass of BaCl2 = 208.3 g/mol
molar mass of BaBr2 = 297.3 g/mol
molar mass of BaI2= 391.3 g/mol
Now calculate the number of moles of BaSO4:
number of moles ;BaSO4 = 0.124 g/ 233.3 g/mol
= 0.00053 moles
Here H2SO4 is present in excess so
BaX2 ............ +..... H2SO4 ------------>BaSO4 + 2HX
0.00053 moles.....0.00053 moles........0.00053 moles
Now we multiply this number of moles BaX2 with all masses of BaX2 and compare with the given value:
0.00053 x 175.3 g/mol (BaF3) = 0.0929 g does not match with the given value of 0.158 g
0.00053 x 208.3 g/mol (BaCl2) = 0.110 g does not match with the given value of 0.158 g
0.00053 x 297.3 g/mol (BaBr2) = 0.157 g is closet with the given value 0.158 g
Hence the formula of BaX2 is BaBr2 (Barium bromide).
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