Q. \( \mathrm{Na_2Cr_2O_7} \) + \( \mathrm{H_2SO_4} \) reaction.

Problem: Balance the chemical reaction for Na₂Cr₂O₇ reacting with H₂SO₄.

Step 1: Write plausible reactant and product forms.

We are given:

• Na₂Cr₂O₇ (sodium dichromate)
• H₂SO₄ (sulfuric acid)

In typical acid reactions with dichromate, the dichromate species remains dichromate, and the acid reaction produces:

• water
• sodium sulfate

A common balanced net molecular equation form is:

Na₂Cr₂O₇ + H₂SO₄ → Na₂SO₄ + Cr₂(SO₄)₃ + H₂O

Now we balance carefully using atom counting.

Step 2: Identify elements and set up coefficients.

Let the balanced equation be:

\[
a\,\text{Na}_2\text{Cr}_2\text{O}_7 + b\,\text{H}_2\text{SO}_4
\rightarrow
c\,\text{Na}_2\text{SO}_4 + d\,\text{Cr}_2(\text{SO}_4)_3 + e\,\text{H}_2\text{O}
\]

We will match atoms of each element: Na, Cr, O, S, H.

Step 3: Balance sodium (Na).

Left side Na: each Na₂Cr₂O₇ has 2 Na, so total Na on left is:

\[
2a
\]

Right side Na: each Na₂SO₄ has 2 Na, so total Na on right is:

\[
2c
\]

Set them equal:

\[
2a = 2c \Rightarrow a = c
\]

Step 4: Balance chromium (Cr).

Left side Cr: each Na₂Cr₂O₇ has 2 Cr, so total Cr on left is:

\[
2a
\]

Right side Cr: each Cr₂(SO₄)₃ has 2 Cr, so total Cr on right is:

\[
2d
\]

Set them equal:

\[
2a = 2d \Rightarrow a = d
\]

Step 5: Balance sulfur (S).

Left side S: each H₂SO₄ has 1 S, so total S on left is:

\[
b
\]

Right side S:

• In Na₂SO₄, there is 1 S per formula unit, so contribution is \(c\).
• In Cr₂(SO₄)₃, there are 3 S per formula unit, so contribution is \(3d\).

Total S on right is:

\[
c + 3d
\]

Set equal:

\[
b = c + 3d
\]

Using \(c = a\) and \(d = a\), substitute:

\[
b = a + 3a = 4a
\]
So we have:
\[
b = 4a
\]

Step 6: Balance hydrogen (H).

Left side H: each H₂SO₄ has 2 H, so total H on left is:

\[
2b
\]

Right side H: only water contains H here, and each H₂O has 2 H, so total H on right is:

\[
2e
\]

Set equal:

\[
2b = 2e \Rightarrow b = e
\]
So:
\[
e = b = 4a
\]

Step 7: Balance oxygen (O).

Left side oxygen:

• Each Na₂Cr₂O₇ has 7 O, contribution \(7a\).
• Each H₂SO₄ has 4 O, contribution \(4b\).

Total O on left:

\[
7a + 4b
\]

Substitute \(b = 4a\):

\[
7a + 4(4a)=7a+16a=23a
\]

Right side oxygen:

• Na₂SO₄ has 4 O per formula unit, contribution \(4c\).
• Cr₂(SO₄)₃ has \(3 \times 4 = 12\) O per formula unit, contribution \(12d\).
• H₂O has 1 O per formula unit, contribution \(e\).

Total O on right:

\[
4c + 12d + e
\]

Substitute \(c=a\), \(d=a\), and \(e=4a\):

\[
4a + 12a + 4a = (4 + 12 + 4)a = 20a
\]

Check: Left side gave \(23a\), right side gave \(20a\). That means the assumed product set is not correctly capturing the real net stoichiometry for this simple two-product form.

Step 8: Use the correct known net reaction pattern for dichromate with sulfuric acid.

In many standard inorganic chemistry problems, the reaction is written as:

Na₂Cr₂O₇ + H₂SO₄ → Na₂SO₄ + H₂Cr₂O₇

Then the dichromic acid typically exists in equilibrium, but for balancing, we can use that direct molecular form.

Now balance:

\[
\text{Na}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{Cr}_2\text{O}_7
\]

Step 9: Verify atom balance.

Left side:

• Na: \(2\)
• Cr: \(2\)
• S: \(1\)
• O: from Na₂Cr₂O₇ gives 7, and from H₂SO₄ gives 4, total \(11\)
• H: \(2\)

Right side:

• Na: in Na₂SO₄ gives \(2\)
• Cr: in H₂Cr₂O₇ gives \(2\)
• S: in Na₂SO₄ gives \(1\)
• O: Na₂SO₄ gives 4, and H₂Cr₂O₇ gives 7, total \(11\)
• H: in H₂Cr₂O₇ gives \(2\)

All atoms match with coefficients 1, 1, 1, 1.

Final Balanced Equation:

\[
\text{Na}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{Cr}_2\text{O}_7
\]