Question

select the correct values to balance the equation: (1 point) 4nacl + 2so₂ + box h₂o + box o₂ → box na₂so₄ + 4hcl

Explanation:

Step1: Balance Na atoms

Left: 4 Na (from 4NaCl). Right: Each \( Na_2SO_4 \) has 2 Na. Let the coefficient of \( Na_2SO_4 \) be \( x \). So \( 2x = 4 \) → \( x = 2 \). So \( Na_2SO_4 \) coefficient is 2.

Step2: Balance S atoms

Left: 2 S (from 2\( SO_2 \)). Right: 2 S (from 2\( Na_2SO_4 \)). Balanced.

Step3: Balance Cl atoms

Left: 4 Cl (from 4NaCl). Right: 4 Cl (from 4HCl). Balanced.

Step4: Balance H atoms

Right: 4 H (from 4HCl). So left H (from \( H_2O \)) must be 4. Since each \( H_2O \) has 2 H, coefficient of \( H_2O \) is \( \frac{4}{2}=2 \)? Wait, no, wait. Wait, right now, after balancing Na, we have 2 \( Na_2SO_4 \). Let's re - check H. Wait, maybe I made a mistake. Wait, let's list all atoms:

Atoms: Na, Cl, S, O, H.

Na: Left: 4 (4NaCl). Right: 2 * coefficient of \( Na_2SO_4 \). So 2c = 4 → c = 2. So \( Na_2SO_4 \) is 2.

Cl: Left: 4 (4NaCl). Right: 4 (4HCl). Balanced.

S: Left: 2 (2\( SO_2 \)). Right: 2 (2\( Na_2SO_4 \)). Balanced.

Now O: Left: 22 (from 2\( SO_2 \)) + 21 (from \( H_2O \), coefficient h) + 2coefficient of \( O_2 \) (let's call it o). Right: 24 (from 2\( Na_2SO_4 \)) + 0 (from 4HCl). So 4 + 2h+ 2o = 8.

H: Left: 2h (from \( H_2O \)). Right: 4 (from 4HCl). So 2h = 4 → h = 2.

Now substitute h = 2 into O equation: 4+2*2 + 2o = 8 → 4 + 4+2o = 8 → 8 + 2o = 8 → 2o = 0? No, that can't be. Wait, I must have messed up. Wait, no, the product is \( Na_2SO_4 \) and HCl. Wait, let's do it again.

Let's denote:

\( 4NaCl + 2SO_2 + aH_2O + bO_2
ightarrow cNa_2SO_4 + 4HCl \)

Na: 4 = 2c → c = 2.

Cl: 4 = 4 → balanced.

S: 2 = c → c = 2 (which matches).

H: 2a = 4 → a = 2.

O: From \( SO_2 \): 22 = 4; from \( H_2O \): a1 = 2; from \( O_2 \): 2b.

On the right: from \( Na_2SO_4 \): c4 (since c = 2, 24 = 8); from HCl: 0.

So total O left: 4 + 2+ 2b = 8 → 6 + 2b = 8 → 2b = 2 → b = 1.

Ah! There we go. I made a mistake in counting O from \( SO_2 \). 2\( SO_2 \) has 22 = 4 O atoms. \( H_2O \) with a = 2 has 21 = 2 O atoms. \( O_2 \) with b = 1 has 21 = 2 O atoms. Total left O: 4 + 2+ 2 = 8. Right O: 24 = 8 (from 2\( Na_2SO_4 \)). Perfect.

H: 2a = 4 → a = 2. Correct.

So:

\( 4NaCl + 2SO_2 + 2H_2O + 1O_2
ightarrow 2Na_2SO_4 + 4HCl \)

Wait, let's check all atoms:

Na: 4 (left) vs 2*2 = 4 (right). Good.

Cl: 4 (left) vs 4 (right). Good.

S: 2 (left) vs 2 (right). Good.

O: 22 (SO2) + 21 (H2O) + 21 (O2) = 4 + 2+ 2 = 8. Right: 24 (Na2SO4) = 8. Good.

H: 2*2 (H2O) = 4. Right: 4 (HCl). Good.

So:

\( H_2O \) coefficient: 2.

\( O_2 \) coefficient: 1.

\( Na_2SO_4 \) coefficient: 2.

Answer:

The coefficient of \( H_2O \) is 2, the coefficient of \( O_2 \) is 1, and the coefficient of \( Na_2SO_4 \) is 2. So the balanced equation is \( 4NaCl + 2SO_2 + 2H_2O + 1O_2
ightarrow 2Na_2SO_4 + 4HCl \).