What will be the value of 'A+B+C' if the following reaction is balanced?

${AFeCl}_{3} (aq) + BMgO (s) 🡪 {Fe}_{2} O_{3} (s) + {CMgCl}_{2} (aq)$

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Solution

To find the value of A,B and C, we need to do the balancing of the given equation.

Number of Fe atoms on the reactant side: 1
Number of Fe atoms on the product side: 2

So we will multiply 2 on the left side to $FeCl 3$ . Hence it will become;

$2 {FeCl}_{3} (aq) + MgO (s) 🡪 {Fe}_{2} O_{3} (s) + {MgCl}_{2} (aq)$

Number of Cl atoms on the reactant side: 6 (2 X 3)
Number of Cl atoms on the product side: 2

So we will multiply 3 on the right side to $M g C l_{2}$ . Hence it will become;

$2 F e C l_{3} (a q) + M g O (s) 🡪 F e_{2} O_{3} (s) + 3 M g C l_{2} (a q)$

Now, the number of Mg atoms are not balanced. In order to balance it, we will multiply 3 to MgO.

$2 {FeCl}_{3} (aq) + 3 MgO (s) 🡪 {Fe}_{2} O_{3} (s) + 3 {MgCl}_{2} (aq)$

Hence the equation is balanced and we also identified the value of A, B, and C, which are 2, 3, and 3 respectively.
So A+B+C =8.

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Similar questions

A+B+C

{AFeCl}_{3} (aq) + BMgO (s) \to {Fe}_{2} O_{3} (s) + {CMgCl}_{2} (aq)

A F e C l_{3} + B M g O \to F e_{2} O_{3} + C M g C l_{2}

F e_{2} O_{3 (s)} + 3 H_{2} O_{(l)}

△ \int H^{0} (H_{2} O, I)

- 285.83 k J m o l^{- 1}

△ \int H^{0} (F e_{2} O_{3}, S)

- 824.2 k J m o l^{- 1}

{xFeCl}_{3} + yMgO \to {Fe}_{2} O_{3} + {zMgCl}_{2}

What will be a, b and c for the following reaction to be balanced?

$a F e C l_{3} + b M g O \to c F e_{2} O_{3} + d M g C l_{2}$

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