Consider the redox reaction below. 2 upper A l (s) plus 6 upper H C l (a q) right arrow 2 upper Al upper C l subscript 3 (a q) plus 3 upper H subscript 2 (g). Which statement correctly describes a half-reaction that is taking place? Hydrogen is oxidized from +1 to 0. Chlorine is reduced from –1 to 0. Aluminum is oxidized from 0 to +3. Hydrogen is reduced from 0 to –1.

Aluminum is oxidized from 0 to +3

Explanation:

In order to make the question clearer, let me give the reaction equation in terms of chemical symbols. It will aid your understanding of the question posed.

2Al(s) + 6HCl(aq) > 2AlCl3(aq) + 3H2(g)

You have to recall that oxidation has to do with increase in oxidation number. A specie is said to be oxidized when the said specie experiences an increase in oxidation number. A reduction on the other hand occurs when a specie undergoes a decrease in oxidation number.

Since the oxidation number of an uncombined element is zero, aluminum has an oxidation number of zero on the left hand side. Over on the right hand side, the oxidation number of aluminium increases to +3. Hence aluminum was oxidized from left to right.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

explanation: