Chemistry, asked by qwer6174, 1 year ago

How many mL of 0.125 M Cr3+ must be reacted with 12 mL of 0.2 M MnO4 -, if the redox products are CrO7 2- and Mn2+ ?

Answers

Answered by BarrettArcher
0

Answer : 19.2 ml of Cr^{+3} reacted with MnO^-_4.

Solution : Given,

Molarity of Cr^{+3} = 0.125 mole/L

Molarity of MnO^-_4 = 0.2 mole/L

Volume of MnO^-_4 = 12 ml = 0.012 L     (1 L = 1000 ml)

The given chemical redox reaction is,

Cr^{3+}+MnO^-_4\rightarrow CrO^{2-}_4+Mn^{2+}

From the given reaction, we conclude that the 1 mole of Cr^{+3} react with the 1 mole of MnO^-_4.

Formula used : M_1V_1=M_2V_2

where,

M_1 = Molarity of Cr^{+3}

V_1 = Volume of Cr^{+3}

M_2 = Molarity of MnO^-_4

V_2 = Volume of MnO^-_4

Now put all the given values in above formula, we get the volume of Cr^{+3}.

0.125mole/L\times V_1=0.2mole/L\times 0.012L\\V_1=0.0192L=19.2ml

Therefore, 19.2 ml of Cr^{+3} reacted with MnO^-_4.

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