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

$$\mathrm{Balance}\mathrm{the}\mathrm{equation}$$$${\mathrm{K}}_2{\mathrm{CrO}}_4+\mathrm{HCl}\to {\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+\mathrm{KCl}+{\mathrm{H}}_2\mathrm{O}$$

Answered by Joel577 last updated on 17/May/17

$${\mathrm{aK}}_2{\mathrm{CrO}}_4+\mathrm{bHCl}\to {\mathrm{cK}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+\mathrm{dKCl}+{\mathrm{eH}}_2\mathrm{O}$$$$\mathrm{In}\mathrm{left}\mathrm{side}\mathrm{In}\mathrm{right}\mathrm{side}$$$$\mathrm{K}=2\mathrm{a}\mathrm{K}=2\mathrm{c}+\mathrm{d}$$$$\mathrm{Cr}=\mathrm{a}\mathrm{Cr}=2\mathrm{c}$$$$\mathrm{O}=4\mathrm{a}\mathrm{O}=7\mathrm{c}+\mathrm{e}$$$$\mathrm{H}=\mathrm{b}\mathrm{H}=2\mathrm{e}$$$$\mathrm{Cl}=\mathrm{b}\mathrm{Cl}=\mathrm{d}$$$$$$$$\mathrm{Assume}\mathrm{that}\mathrm{d}=1$$$$\bullet \mathrm{b}=\mathrm{d}=1$$$$\bullet \mathrm{b}=2\mathrm{e}\to \mathrm{e}=\frac{1}{2}$$$$\bullet 2\mathrm{a}=2\mathrm{c}+\mathrm{d}\to 4\mathrm{a}-4\mathrm{c}=2$$$$4\mathrm{a}=7\mathrm{c}+\mathrm{e}\to \underset{\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}}{4\mathrm{a}-7\mathrm{c}=\frac{1}{2}}$$$$3\mathrm{c}=\frac{3}{2}\to \mathrm{c}=\frac{1}{2}$$$$\bullet \mathrm{a}=1$$$$$$$$\therefore {\mathrm{K}}_2{\mathrm{CrO}}_4+\mathrm{HCl}\to \frac{1}{2}{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+\mathrm{KCl}+\frac{1}{2}{\mathrm{H}}_2\mathrm{O}$$$${2\mathrm{K}}_2{\mathrm{CrO}}_4+2\mathrm{HCl}\to {\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+2\mathrm{KCl}+{\mathrm{H}}_2\mathrm{O}$$

Commented by Tinkutara last updated on 15/Jul/17

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

Answered by Tinkutara last updated on 15/Jul/17

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