How do you balance $A g N O_{3} + M g C l_{2} \to A g C l + M {g (N O_{3})}_{2}$ $AgNO_3 +MgCl_2 → AgCl + Mg(NO_3)_2$ ?

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How do you balance $A g N O_{3} + M g C l_{2} \to A g C l + M {g (N O_{3})}_{2}$ $AgNO_3 +MgCl_2 → AgCl + Mg(NO_3)_2$ ?

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Answer 1 · 2017-03-01T18:21:56

$2 {AgNO}_{3} + MgC l_{2} \to 2 AgC l + {Mg(NO}_{3})_{2}$ $2"AgNO"_3 +"MgC"l_2 → 2"AgC"l + "Mg(NO"_3")"_2$

Explanation:

On the RHS, we have two nitrate ions. Therefore, we need $2$ $2$ mols of ${AgNO}_{3}$ $"AgNO"_3$ to balance them. We now have two mols of ${Ag}^{+}$ $"Ag"^+$ on the LHS but only one mol on the RHS. Thus we also need $2$ $2$ mols $AgC l$ $"AgC"l$ .

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How do you balance $A g N O_{3} + M g C l_{2} \to A g C l + M {g (N O_{3})}_{2}$ $AgNO_3 +MgCl_2 → AgCl + Mg(NO_3)_2$ ?

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How do you balance AgNO_3 +MgCl_2 → AgCl + Mg(NO_3)_2AgNO3+MgCl2→AgCl+Mg(NO3)2AgNO_3 +MgCl_2 → AgCl + Mg(NO_3)_2?

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How do you balance $A g N O_{3} + M g C l_{2} \to A g C l + M {g (N O_{3})}_{2}$ $AgNO_3 +MgCl_2 → AgCl + Mg(NO_3)_2$ ?