Question

1. zns + alp → zn₃p₂ + al₂s₃

Explanation:

Step 1: Balance Zn

On the left, we have 1 Zn (in ZnS), and on the right, we have 3 Zn (in \( Zn_3P_2 \)). So we put a coefficient of 3 in front of ZnS:
\( 3ZnS + AlP ightarrow Zn_3P_2 + Al_2S_3 \)

Step 2: Balance S

Now, with 3 ZnS, we have 3 S on the left. On the right, in \( Al_2S_3 \), we have 3 S. So S is balanced for now.

Step 3: Balance Al and P

On the right, we have \( Zn_3P_2 \) (2 P) and \( Al_2S_3 \) (2 Al). On the left, we have AlP (1 Al and 1 P). To balance P: we need 2 P, so coefficient 2 for AlP? Wait, no—wait, \( Zn_3P_2 \) has 2 P, so we need 2 P from AlP. But also, \( Al_2S_3 \) has 2 Al, so we need 2 Al from AlP. Wait, let's re-examine.

Wait, after balancing Zn (3 ZnS), let's look at S: 3 S in 3 ZnS, so \( Al_2S_3 \) needs 3 S, which it has (since \( Al_2S_3 \) has 3 S). Now, for Al: \( Al_2S_3 \) has 2 Al, so we need 2 Al from AlP. For P: \( Zn_3P_2 \) has 2 P, so we need 2 P from AlP. So AlP needs a coefficient of 2? Wait, no—wait, AlP has 1 Al and 1 P. So to get 2 Al and 2 P, we need 2 AlP? Wait, no, 2 AlP would give 2 Al and 2 P. Let's check:

Left: 3 ZnS + 2 AlP
Right: \( Zn_3P_2 + Al_2S_3 \)

Wait, no, \( Al_2S_3 \) has 2 Al, so 2 Al from 2 AlP (since each AlP has 1 Al). \( Zn_3P_2 \) has 2 P, so 2 P from 2 AlP (each AlP has 1 P). Now check S: 3 ZnS has 3 S, \( Al_2S_3 \) has 3 S. Good. Zn: 3 ZnS has 3 Zn, \( Zn_3P_2 \) has 3 Zn. Good. Al: 2 AlP has 2 Al, \( Al_2S_3 \) has 2 Al. Good. P: 2 AlP has 2 P, \( Zn_3P_2 \) has 2 P. Good. Wait, but wait, let's write the coefficients:

Wait, no—wait, if we put 3 ZnS, 2 AlP, then:

Left: 3 ZnS + 2 AlP
Right: \( Zn_3P_2 + Al_2S_3 \)

Wait, but \( Al_2S_3 \) has 3 S, and 3 ZnS has 3 S. \( Zn_3P_2 \) has 3 Zn, 3 ZnS has 3 Zn. 2 AlP has 2 Al and 2 P, \( Al_2S_3 \) has 2 Al, \( Zn_3P_2 \) has 2 P. Wait, but that doesn't balance. Wait, no—wait, I made a mistake. Let's do it properly.

Wait, let's list the atoms:

- Zn: left (ZnS) = 1, right (\( Zn_3P_2 \)) = 3 → coefficient 3 for ZnS: \( 3ZnS \)
- S: left (3 ZnS) = 3, right (\( Al_2S_3 \)) = 3 → good.
- Al: right (\( Al_2S_3 \)) = 2, left (AlP) = 1 → need coefficient 2 for AlP? Wait, no, AlP has 1 Al, so to get 2 Al, coefficient 2 for AlP. But then P: left (2 AlP) = 2, right (\( Zn_3P_2 \)) = 2 → good.

Wait, so putting it all together:

\( 3ZnS + 2AlP ightarrow Zn_3P_2 + Al_2S_3 \)

Wait, but let's check all atoms:

- Zn: 3 (left: 3 ZnS) vs 3 (right: \( Zn_3P_2 \)) → good.
- S: 3 (left: 3 ZnS) vs 3 (right: \( Al_2S_3 \)) → good.
- Al: 2 (left: 2 AlP) vs 2 (right: \( Al_2S_3 \)) → good.
- P: 2 (left: 2 AlP) vs 2 (right: \( Zn_3P_2 \)) → good.

Wait, that works? Wait, no, wait—wait, \( Al_2S_3 \) has 3 S, and 3 ZnS has 3 S. Yes. So the balanced equation is:

\( 3ZnS + 2AlP ightarrow Zn_3P_2 + Al_2S_3 \)

Wait, but let me check again. Let's count each atom:

Left:
- Zn: 3 (3 × 1)
- S: 3 (3 × 1)
- Al: 2 (2 × 1)
- P: 2 (2 × 1)

Right:
- Zn: 3 (1 × 3)
- P: 2 (1 × 2)
- Al: 2 (1 × 2)
- S: 3 (1 × 3)

Yes! All atoms are balanced.

Wait, but wait, initially I thought maybe I messed up, but no—this works. So the balanced equation is \( 3ZnS + 2AlP ightarrow Zn_3P_2 + Al_2S_3 \).

Wait, but let me confirm with another approach. Let's use variables. Let the coefficients be \( aZnS + bAlP ightarrow cZn_3P_2 + dAl_2S_3 \).

Balance Zn: \( a = 3c \)
Balance S: \( a = 3d \)
Balance Al: \( b = 2d \)
Balance P: \( b = 2c \)

From Zn: \( a = 3c \)
From S: \( a = 3d \) → so \( 3c = 3d \) → \( c = d \)
From Al: \( b = 2d \)
From P: \( b = 2c \) → but \( c = d \), so \( b = 2c \) and \( b = 2d \) (same thing).

Let’s set \( c = 1 \), then \( a = 3(1) = 3 \), \( d = c = 1 \), \( b = 2(1) = 2 \).

So \( a = 3 \), \( b = 2 \), \( c = 1 \), \( d = 1 \).

Thus, the balanced equation is \( 3ZnS + 2AlP ightarrow Zn_3P_2 + Al_2S_3 \).

Answer:

The balanced chemical equation is \( \boldsymbol{3ZnS + 2AlP
ightarrow Zn_3P_2 + Al_2S_3} \).