Number of moles of K_(2)Cr_(2)O_7 , reduced by one mole of Sn^(2+) ions is 1/x.The value of x is

Number of moles of K_(2)Cr_(2)O_7 , reduced by one mole of Sn^(2+) ion

1.	Number of moles of K_(2)Cr_(2)O_7 , reduced by one mole of Sn^(2+) ions is 1/x.The value of x is
Answer» Solution :`Cr_(2)O_(7)^(-2)+14H^(+) +3SN^(2+) rarr2Cr^(3+) +3Sn^(+4) +7H_(2)O`. Thus ,1 moles of `Sn^(2+)`reduces `1/3` moles of `K_(2)Cr_(2)O_7`