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Question Number 13261 by Tinkutara last updated on 17/May/17

Balance the equation K_2 CrO_4 + HCl → K_2 Cr_2 O_7 + KCl + H_2 O

$$\mathrm{Balance}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{K}_{\mathrm{2}} \mathrm{CrO}_{\mathrm{4}} \:+\:\mathrm{HCl}\:\rightarrow\:\mathrm{K}_{\mathrm{2}} \mathrm{Cr}_{\mathrm{2}} \mathrm{O}_{\mathrm{7}} \:+\:\mathrm{KCl}\:+\:\mathrm{H}_{\mathrm{2}} \mathrm{O} \\ $$

Answered by Joel577 last updated on 17/May/17

aK_2 CrO_4 + bHCl → cK_2 Cr_2 O_7 + dKCl + eH_2 O In left side In right side K = 2a K = 2c + d Cr = a Cr = 2c O = 4a O = 7c + e H = b H = 2e Cl = b Cl = d Assume that d = 1 • b = d = 1 • b = 2e → e = (1/2) • 2a = 2c + d → 4a − 4c = 2 4a = 7c + e → 4a − 7c = (1/2)_(_______________) 3c = (3/2) → c = (1/2) • a = 1 ∴ K_2 CrO_4 + HCl → (1/2)K_2 Cr_2 O_7 + KCl + (1/2)H_2 O 2K_2 CrO_4 + 2HCl → K_2 Cr_2 O_7 + 2KCl + H_2 O

$$\mathrm{aK}_{\mathrm{2}} \mathrm{CrO}_{\mathrm{4}} \:+\:\mathrm{bHCl}\:\rightarrow\:\mathrm{cK}_{\mathrm{2}} \mathrm{Cr}_{\mathrm{2}} \mathrm{O}_{\mathrm{7}} \:+\:\mathrm{dKCl}\:+\:\mathrm{eH}_{\mathrm{2}} \mathrm{O} \\ $$$$\mathrm{In}\:\mathrm{left}\:\mathrm{side}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{In}\:\mathrm{right}\:\mathrm{side} \\ $$$$\:\mathrm{K}\:=\:\mathrm{2a}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{K}\:=\:\mathrm{2c}\:+\:\mathrm{d} \\ $$$$\mathrm{Cr}\:=\:\mathrm{a}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Cr}\:=\:\mathrm{2c}\:\:\:\:\:\: \\ $$$$\:\:\mathrm{O}\:=\:\mathrm{4a}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{O}\:=\:\mathrm{7c}\:+\:\mathrm{e} \\ $$$$\:\:\mathrm{H}\:=\:\mathrm{b}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{H}\:=\:\mathrm{2e} \\ $$$$\:\mathrm{Cl}\:=\:\mathrm{b}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Cl}\:=\:\mathrm{d} \\ $$$$ \\ $$$$\mathrm{Assume}\:\mathrm{that}\:\mathrm{d}\:=\:\mathrm{1} \\ $$$$\bullet\:\mathrm{b}\:=\:\mathrm{d}\:=\:\mathrm{1} \\ $$$$\bullet\:\mathrm{b}\:=\:\mathrm{2e}\:\:\rightarrow\:\:\mathrm{e}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\bullet\:\mathrm{2a}\:=\:\mathrm{2c}\:+\:\mathrm{d}\:\:\rightarrow\:\:\:\mathrm{4a}\:−\:\mathrm{4c}\:=\:\mathrm{2} \\ $$$$\:\:\:\:\mathrm{4a}\:=\:\mathrm{7c}\:+\:\mathrm{e}\:\:\rightarrow\:\:\underset{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} {\mathrm{4a}\:−\:\mathrm{7c}\:=\:\frac{\mathrm{1}}{\mathrm{2}}}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3c}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:\:\rightarrow\:\:\mathrm{c}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\bullet\:\mathrm{a}\:=\:\mathrm{1} \\ $$$$ \\ $$$$\therefore\:\mathrm{K}_{\mathrm{2}} \mathrm{CrO}_{\mathrm{4}} \:+\:\mathrm{HCl}\:\rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{K}_{\mathrm{2}} \mathrm{Cr}_{\mathrm{2}} \mathrm{O}_{\mathrm{7}} \:+\:\mathrm{KCl}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{H}_{\mathrm{2}} \mathrm{O} \\ $$$$\:\:\:\:\:\mathrm{2K}_{\mathrm{2}} \mathrm{CrO}_{\mathrm{4}} \:+\:\mathrm{2HCl}\:\rightarrow\:\mathrm{K}_{\mathrm{2}} \mathrm{Cr}_{\mathrm{2}} \mathrm{O}_{\mathrm{7}} \:+\:\mathrm{2KCl}\:+\:\mathrm{H}_{\mathrm{2}} \mathrm{O} \\ $$

Commented by Tinkutara last updated on 15/Jul/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Answered by Tinkutara last updated on 15/Jul/17

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