Question

k₂so₃(aq) + mncl₂(aq) → mnso₃(s) + kcl(aq)
express your answers as integers separated by commas.

submit request answer

part b

n₂h₄(l) → nh₃(g) + n₂(g)
express your answers as integers separated by commas.

submit request answer

part c

hclo₃(aq) + hbr(aq) → hcl(aq) + br₂(g) + h₂o(l)
express your answers as integers separated by commas.

Explanation:

Part A

Step1: Balance K

On left: 2 K (from \(K_2SO_3\)). On right: 1 K (from \(KCl\)). So put 2 in front of \(KCl\).
Reaction: \(K_2SO_3(aq) + MnCl_2(aq) \to MnSO_3(s) + 2KCl(aq)\)

Step2: Check other elements

Mn: 1 on left (\(MnCl_2\)), 1 on right (\(MnSO_3\)) – balanced.
S: 1 on left (\(K_2SO_3\)), 1 on right (\(MnSO_3\)) – balanced.
Cl: 2 on left (\(MnCl_2\)), 2 on right (2\(KCl\)) – balanced.
O: 3 on left (\(K_2SO_3\)), 3 on right (\(MnSO_3\)) – balanced.

Step1: Balance N

Left: 2 N (from \(N_2H_4\)). Right: 1 N (\(NH_3\)) + 2 N (\(N_2\)) = 3 N. Find LCM of 2 and 3, which is 6. So adjust coefficients:
Let coefficient of \(N_2H_4\) be 3, \(NH_3\) be 4, \(N_2\) be 1 (since 32 = 41 + 2*1? Wait, better way:
Let \(aN_2H_4 \to bNH_3 + cN_2\)
N: 2a = b + 2c
H: 4a = 3b
Let a=3, then H: 12=3b ⇒ b=4. Then N: 6=4 + 2c ⇒ c=1.
Reaction: \(3N_2H_4(l) \to 4NH_3(g) + N_2(g)\)

Step2: Verify

N: 6 (3*2) = 4 + 2 (4+2=6) – balanced.
H: 12 (34) = 12 (43) – balanced.

Step1: Balance Cl and Br (redox)

\(HClO_3\) (Cl from +5 to -1, gain 6 e⁻)
\(HBr\) (Br from -1 to 0, lose 1 e⁻ per Br, so 2 Br⁻ lose 2 e⁻ for \(Br_2\))
To balance electrons, multiply \(HClO_3\) by 1, \(HBr\) by 6 (since 6 e⁻ gained, 6 e⁻ lost: 61 e⁻ lost per 2 Br⁻? Wait, \(Br_2\) has 2 Br, so for each \(Br_2\), 2 e⁻ lost. So to balance 6 e⁻ gain, need 3 \(Br_2\) (32=6 e⁻ lost). So:
\(HClO_3 + 6HBr \to HCl + 3Br_2 + H_2O\)

Step2: Balance H and O

H: Left: 1 + 6 = 7. Right: 1 (HCl) + 2x (H₂O). Let x=3: 1 + 6 = 7. O: Left: 3 (HClO₃), Right: 3 (3H₂O) – balanced.
Reaction: \(HClO_3(aq) + 6HBr(aq) \to HCl(aq) + 3Br_2(g) + 3H_2O(l)\)

Step3: Verify

Cl: 1 on left, 1 on right – balanced.
Br: 6 on left (6HBr), 6 on right (3Br₂) – balanced.
H: 7 on left (1+6), 7 on right (1 + 6) – balanced.
O: 3 on left, 3 on right – balanced.

Answer:

1,1,1,2

Part B