Consider the following equation
`H_(4)P_(2)O_(7) + 2NaOH to Na_(2)H_(2)P_(2)O_(7) + 2H_(2)O`
If 534 gm of `H_(4)P_(2)O_(7)` is reacted with `3.0 xx 10^(24)` formula units of NaOH ,then total number of moles of `H_(2)O` is produced is (`N_(A)` = `6 xx 10^(23)` )

Answer

To solve the problem, we need to determine the total number of moles of water (H₂O) produced when 534 grams of H₄P₂O₇ reacts with 3.0 x 10²⁴ formula units of NaOH, based on the provided chemical equation:

\[ H_4P_2O_7 + 2NaOH \rightarrow Na_2H_2P_2O_7 + 2H_2O \]

### Step 1: Calculate the moles of NaOH
We know that the number of moles can be calculated using the formula:

\[ \text{Moles} = \frac{\text{Number of formula units}}{N_A} \]

Where \( N_A \) (Avogadro's number) is \( 6 \times 10^{23} \).

Given:
- Number of formula units of NaOH = \( 3.0 \times 10^{24} \)

Calculating the moles of NaOH:

\[ \text{Moles of NaOH} = \frac{3.0 \times 10^{24}}{6 \times 10^{23}} = 5 \text{ moles} \]

### Step 2: Calculate the moles of H₄P₂O₇
Next, we need to calculate the moles of H₄P₂O₇ using its molar mass. The molar mass of H₄P₂O₇ is given as 178 g/mol.

Given:
- Mass of H₄P₂O₇ = 534 g

Calculating the moles of H₄P₂O₇:

\[ \text{Moles of H}_4P_2O_7 = \frac{534 \text{ g}}{178 \text{ g/mol}} \approx 3 \text{ moles} \]

### Step 3: Identify the limiting reagent
From the balanced equation, we see that:

- 1 mole of H₄P₂O₇ reacts with 2 moles of NaOH.

For 3 moles of H₄P₂O₇, we would need:

\[ 3 \text{ moles H}_4P_2O_7 \times 2 \text{ moles NaOH/mole H}_4P_2O_7 = 6 \text{ moles NaOH} \]

Since we only have 5 moles of NaOH available, NaOH is the limiting reagent.

### Step 4: Calculate the moles of H₂O produced
According to the balanced equation:

- 2 moles of NaOH produce 2 moles of H₂O.

Thus, 5 moles of NaOH will produce:

\[ 5 \text{ moles NaOH} \times \frac{2 \text{ moles H}_2O}{2 \text{ moles NaOH}} = 5 \text{ moles H}_2O \]

### Final Answer
The total number of moles of H₂O produced is **5 moles**.

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