claculate n factor of hcl in equation K 2 Cr 2 O 7 + 14 HCl = 2 KCl + 2 CrCl 3 + 2 Cl 3 + 7 H 2 O

Question

Arun · Answer

You should write chemical formula correctly, using caps when required, such as...
Kr2Cr2O7 + HCl ==> KCl + CrCl3 + + Cl2 + H2O and balanced it would be ...
14HCl + K2Cr2O7 → 3Cl2 + 2CrCl3 + 2KCl + 7H2O
and n for HCl is 14

Shireen Mujawar · Answer

In balanced chemical reaction, 14 mole of HCl looses 6 mole of electrons So,1 mole of HCl looses 6/14 mole of electrons So, n factor of HCl is 6/14 i.e. 3/7

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claculate n factor of hcl in equation K 2 Cr 2 O 7 + 14 HCl = 2 KCl + 2 CrCl 3 + 2 Cl 3 + 7 H 2 O

claculate n factor of hcl in equation
K₂Cr₂O₇ + 14 HCl = 2 KCl + 2 CrCl₃ + 2 Cl₃ + 7 H₂O

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claculate n factor of hcl in equation K 2 Cr 2 O 7 + 14 HCl = 2 KCl + 2 CrCl 3 + 2 Cl 3 + 7 H 2 O

claculate n factor of hcl in equationK2Cr2O7 + 14 HCl = 2 KCl + 2 CrCl3 + 2 Cl3 + 7 H2O

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claculate n factor of hcl in equation
K₂Cr₂O₇ + 14 HCl = 2 KCl + 2 CrCl₃ + 2 Cl₃ + 7 H₂O