$KMn{O_4}$ reacts with oxalic acid according to the equation, $2MnO_4^ - + 5{C_2}O_4^{2 - } + 16{H^ + } o 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O$, here 20 ml of 0.1 M $KMn{O_4}$ is equivalent to

Question

$KMn{O_4}$ reacts with oxalic acid according to the equation, $2MnO_4^ -  + 5{C_2}O_4^{2 - } + 16{H^ + } 	o 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O$, here 20 ml of 0.1 M $KMn{O_4}$ is equivalent to

Accepted Answer

50 ml of 0.1 M ${H_2}{C_2}{O_4}$

Answer

20 ml of 0.5 M ${H_2}{C_2}{O_4}$

Answer

50 ml of 0.5 M ${H_2}{C_2}{O_4}$

Answer

20 ml of 0.1 M ${H_2}{C_2}{O_4}$

chemical-aritmatic

Question: $KMn{O_4}$ reacts with oxalic acid according to the equation, $2MnO_4^ - + 5{C_2}O_4^{2 - } + 16{H^ + } \to 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O$, here 20 ml of 0.1 M $KMn{O_4}$ is equivalent to

chemical-aritmatic

Question: $KMn{O_4}$ reacts with oxalic acid according to the equation, $2MnO_4^ - + 5{C_2}O_4^{2 - } + 16{H^ + } \to 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O$, here 20 ml of 0.1 M $KMn{O_4}$ is equivalent to

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