Find the mass of AlCl3 that is produced when 3.58 moles of Al2O3 react with excess HCl according to the following equation.

Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(l)

Moles of Al2O3, m = 3.58 moles.

We need to find the mass of AlCl3 produced.

Solution :

1 mole of Al2O3 produce 2 moles of AlCl3.

So, 3.58 moles of AlCl3 produce 7.16 moles of AlCl3.

Mass of AlCl3 produced is :

m = moles × molecular mass of AlCl3

m = 7.16 × 133.34 g

m = 954.714 g

Therefore, mass of AlCl3 produced is 954.714 g.

Hence, this is the required solution.

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answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.