Question

balancing equations
balance the following chemical equations.
1.
fe +
h₂so₄ →
fe₂(so₄)₃ +
h₂
2.
c₂h₆ +
o₂ →
h₂o +
co₂
3.
koh +
h₃po₄ →
k₃po₄ +
h₂o
4.
sno₂ +
h₂ →
sn +
h₂o
5.
nh₃ +
o₂ →
no +
h₂o
6.
kno₃ +
h₂co₃ →
k₂co₃ +
hno₃
7.
b₂br₆ +
hno₃ →
b(no₃)₃ +
hbr
8.
bf₃ +
li₂so₃ →
b₂(so₃)₃ +
lif
9.
(nh₄)₃po₄ +
pb(no₃)₄ →
pb₃(po₄)₄ +
nh₄no₃
10.
secl₆ +
o₂ →
seo₂ +
cl₂

Explanation:

Step1: Identify atoms to balance

List atoms: Se, Cl, O. Start with Se (1 on left, 1 on right – balanced). Cl: 4 on left (SeCl₄), 2 on right (Cl₂). O: 2 on left (O₂), 2 on right (SeO₂ – balanced for Se, O in SeO₂ is 2).

Step2: Balance Cl

To balance Cl: SeCl₄ has 4 Cl, Cl₂ has 2. So coefficient for Cl₂: 2 (4 Cl total). Now Cl: 4 left, 4 right (2×2).

Step3: Balance O

O: left O₂ (2 O per molecule), right SeO₂ (2 O). Wait, SeO₂ has 2 O, and we have 1 SeO₂. O₂: let's check. Wait, SeCl₄ + O₂ → SeO₂ + Cl₂. Se: 1 left, 1 right. Cl: 4 left, need 2 Cl₂ (2×2=4). O: 2 left (O₂), 2 right (SeO₂). Wait, no – SeO₂ has 2 O, so O₂: 1 molecule (2 O) gives 2 O in SeO₂. Wait, no, let's re-express:

Let coefficients be a SeCl₄ + b O₂ → c SeO₂ + d Cl₂.

Se: a = c

Cl: 4a = 2d → d = 2a

O: 2b = 2c → b = c

Let a = 1, then c = 1, d = 2, b = 1. So equation: 1 SeCl₄ + 1 O₂ → 1 SeO₂ + 2 Cl₂. Wait, but check O: left O₂ (2 O), right SeO₂ (2 O) – balanced. Cl: 4 left, 4 right (2×2). Se: 1 left, 1 right. Wait, but the original formula is SeCl₄? Wait, the problem says SeCl₄? Wait, the user's 10th problem: SeCl₄ + O₂ → SeO₂ + Cl₂? Wait, the image shows SeCl₄? Wait, the handwritten part: "SeCl₄" (maybe typo, but assuming SeCl₄). Wait, no, maybe SeCl₅? Wait, no, the user's 10th: "SeCl₅" + O₂ → SeO₂ + Cl₂? Wait, no, the original: "SeCl₄" (maybe). Wait, let's check again.

Wait, the 10th problem: SeCl₄ (assuming) + O₂ → SeO₂ + Cl₂.

Wait, Se: 1 left, 1 right. Cl: 4 left, 2 right. O: 2 left, 2 right.

So balance Cl: 4 Cl left, so Cl₂ needs coefficient 2 (2×2=4). Then O: 2 left (O₂) → 2 right (SeO₂, which has 2 O). So coefficients:

SeCl₄: 1, O₂: 1, SeO₂: 1, Cl₂: 2. Wait, but O: 2 left (O₂) and 2 right (SeO₂) – balanced. Cl: 4 left, 4 right (2×2). Se: 1 left, 1 right. So the balanced equation is:

$\boldsymbol{1}$ SeCl₄ + $\boldsymbol{1}$ O₂ → $\boldsymbol{1}$ SeO₂ + $\boldsymbol{2}$ Cl₂

Wait, but maybe the original is SeCl₅? No, the user's image: "SeCl₅" + O₂ → SeO₂ + Cl₂? Wait, no, the 10th: "SeCl₄" (maybe a typo, but let's proceed with SeCl₄). Wait, no, maybe the problem is SeCl₄? Wait, the user's 10th: "SeCl₄" + O₂ → SeO₂ + Cl₂. So balancing:

Se: 1 = 1

Cl: 4 = 2d → d=2

O: 2b = 2c → b=c

Let a=1, c=1, b=1, d=2. So coefficients: 1, 1, 1, 2.

Wait, but let's check again. If it's SeCl₄:

1 SeCl₄ + 1 O₂ → 1 SeO₂ + 2 Cl₂

Atoms:

Se: 1 vs 1 – good.

Cl: 4 vs 2×2=4 – good.

O: 2 vs 2 – good.

Yes, that works.

Answer:

1, 1, 1, 2 (coefficients for SeCl₄, O₂, SeO₂, Cl₂ respectively)