What is the balanced equation when aluminum reacts with copper (ii) sulfate? i don't know how to do this
1 answer:
Here Cu (II) reduced to Cu (0). Al would be strong reducing agent to do so. Now ... Compare Al and Cu reduction potentials. Al3+(aq) + 3 e– ----> Al(s) -1.66
Cu(OH)2(s) + 2 e– ----> Cu(s) + 2 OH–(aq) -0.36. It implies that reduction potential of Cu is more than Al hence this reaction is not feasible bcuz copper is a reducing agent ( oxidize itself or reduces others). But still you can balance...practically this reaction seems impossible bcuz of above reason.
Cu(OH)2(s) + 2 e– ----> Cu(s) + 2 OH–(aq) -0.36. It implies that reduction potential of Cu is more than Al hence this reaction is not feasible bcuz copper is a reducing agent ( oxidize itself or reduces others). But still you can balance...practically this reaction seems impossible bcuz of above reason.
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Answer : The concentration of NOBr after 95 s is, 0.013 M
Explanation :
The integrated rate law equation for second order reaction follows:
where,
k = rate constant =
t = time taken = 95 s
[A] = concentration of substance after time 't' = ?
= Initial concentration = 0.86 M
Now put all the given values in above equation, we get:
[A] = 0.013 M
Hence, the concentration of NOBr after 95 s is, 0.013 M
The percent yield of the reaction : 89.14%
<h3>Further explanation</h3>Reaction of Ammonia and Oxygen in a lab :
<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>
mass NH₃ = 80 g
mol NH₃ (MW=17 g/mol):
mass O₂ = 120 g
mol O₂(MW=32 g/mol) :
Mol ratio of reactants(to find limiting reatants) :
mol of H₂O based on O₂ as limiting reactants :
mol H₂O :
mass H₂O :
4.5 x 18 g/mol = 81 g
The percent yield :