How many grams of `H_(3)PO_(4)` is required to completely neutralize 120 g of `NaOH`

Options

A. 49
B. 98
C. 196
D. 9.8

Explanation

To determine how many grams of `H₃PO₄` (phosphoric acid) are required to completely neutralize 120 g of `NaOH` (sodium hydroxide), we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between phosphoric acid and sodium hydroxide can be represented as follows:

\[ H_3PO_4 + 3 NaOH \rightarrow Na_3PO_4 + 3 H_2O \]

This equation shows that one mole of `H₃PO₄` reacts with three moles of `NaOH`.

### Step 2: Calculate the number of moles of `NaOH`
To find the number of moles of `NaOH`, we use the formula:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

The molar mass of `NaOH` is approximately 40 g/mol. Therefore, for 120 g of `NaOH`:

\[ \text{Number of moles of NaOH} = \frac{120 \, \text{g}}{40 \, \text{g/mol}} = 3 \, \text{moles} \]

### Step 3: Determine the moles of `H₃PO₄` required
From the balanced equation, we see that 1 mole of `H₃PO₄` is needed for every 3 moles of `NaOH`. Therefore, the number of moles of `H₃PO₄` required is:

\[ \text{Number of moles of H₃PO₄} = \frac{3 \, \text{moles of NaOH}}{3} = 1 \, \text{mole} \]

### Step 4: Calculate the mass of `H₃PO₄` required
Now, we need to find the mass of 1 mole of `H₃PO₄`. The molar mass of `H₃PO₄` is approximately 98 g/mol. Thus, the mass of `H₃PO₄` required is:

\[ \text{Mass of H₃PO₄} = 1 \, \text{mole} \times 98 \, \text{g/mol} = 98 \, \text{g} \]

### Final Answer
To completely neutralize 120 g of `NaOH`, 98 g of `H₃PO₄` is required.

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