Question

1. given the reaction.

_k₂cr₂o₇ + _hcl → _kcl + _crcl₃ + _cl₂ + _h₂o

when the reaction is completely balanced using smallest whole numbers, the coefficient of cl₂ will be

a. 1
b. 2
c. 3
d. 4

Explanation:

Step1: Assign oxidation numbers

• In \(K_2Cr_2O_7\), \(Cr\) has an oxidation number of \(+6\). In \(CrCl_3\), \(Cr\) has an oxidation number of \(+3\). So each \(Cr\) atom gains \(3\) electrons. Since there are \(2\) \(Cr\) atoms in \(K_2Cr_2O_7\), total electrons gained by \(Cr\) is \(2\times3 = 6\).
• In \(HCl\), \(Cl\) has an oxidation number of \(- 1\). In \(Cl_2\), \(Cl\) has an oxidation number of \(0\). Each \(Cl_2\) molecule loses \(2\) electrons (as \(2\) \(Cl\) atoms each lose \(1\) electron).

Step2: Balance electrons

• To balance electrons, we need \(3\) \(Cl_2\) molecules (since \(3\times2=6\) electrons lost) for every \(1\) \(K_2Cr_2O_7\) (which gains \(6\) electrons).

Step3: Balance other atoms

• Balance \(K\) atoms: \(2\) \(K\) in \(K_2Cr_2O_7\), so \(2\) \(KCl\).
• Balance \(Cr\) atoms: \(2\) \(Cr\) in \(K_2Cr_2O_7\), so \(2\) \(CrCl_3\).
• Balance \(Cl\) atoms: Total \(Cl\) on right - hand side from \(KCl\) and \(CrCl_3\) and \(Cl_2\) is \(2 + 6+6 = 14\). So \(14\) \(HCl\).
• Balance \(O\) and \(H\) atoms: \(7\) \(O\) in \(K_2Cr_2O_7\), so \(7\) \(H_2O\).

The balanced equation is \(K_2Cr_2O_7+14HCl = 2KCl + 2CrCl_3+3Cl_2+7H_2O\)

Answer:

C. 3