A containing ammonium chloride was warmed with 150.0 mL of 1.00mol/L sodium hydroxide the ammonia had been driven off, the excess sodium hydroxide required 50.00 mL of 0.150mol/L sulfuric acid for neutralizatior 1. What mass of ammonium chloride did the sample contain?

To determine the mass of ammonium chloride (NH₄Cl) in the sample, we need to follow these steps:<br /><br />1. Calculate the moles of sodium hydroxide (NaOH) used.<br />2. Determine the moles of sulfuric acid (H₂SO₄) used for neutralization.<br />3. Use stoichiometry to find the moles of NH₄Cl.<br />4. Finally, calculate the mass of NH₄Cl.<br /><br />### Step 1: Calculate the moles of NaOH<br /><br />Given:<br />- Volume of NaOH solution = 150.0 mL = 0.150 L<br />- Concentration of NaOH solution = 1.00 mol/L<br /><br />\[ \text{Moles of NaOH} = \text{Volume (L)} \times \text{Concentration (mol/L)} \]<br />\[ \text{Moles of NaOH} = 0.150 \, \text{L} \times 1.00 \, \text{mol/L} = 0.150 \, \text{mol} \]<br /><br />### Step 2: Determine the moles of H₂SO₄ used for neutralization<br /><br />Given:<br />- Volume of H₂SO₄ solution = 50.00 mL = 0.050 L<br />- Concentration of H₂SO₄ solution = 0.150 mol/L<br /><br />\[ \text{Moles of H₂SO₄} = \text{Volume (L)} \times \text{Concentration (mol/L)} \]<br />\[ \text{Moles of H₂SO₄} = 0.050 \, \text{L} \times 0.150 \, \text{mol/L} = 0.0075 \, \text{mol} \]<br /><br />### Step 3: Use stoichiometry to find the moles of NH₄Cl<br /><br />The balanced chemical equation for the neutralization reaction is:<br /><br />\[ 2 \text{NH₄Cl} + 2 \text{NaOH} \rightarrow 2 \text{NH₃} + 2 \text{H₂O} + 2 \text{NaCl} \]<br /><br />From the equation, 2 moles of NH₄Cl react with 2 moles of NaOH. Therefore, the moles of NH₄Cl are equal to the moles of NaOH used:<br /><br />\[ \text{Moles of NH₄Cl} = 0.150 \, \text{mol} \]<br /><br />### Step 4: Calculate the mass of NH₄Cl<br /><br />The molar mass of NH₄Cl is calculated as follows:<br />- N: 14.01 g/mol<br />- H: 1.01 g/mol (×4 = 4.04 g/mol)<br />- Cl: 35.45 g/mol<br /><br />\[ \text{Molar mass of NH₄Cl} = 14.01 + 4.04 + 35.45 = 53.50 \, \text{g/mol} \]<br /><br />Now, calculate the mass of NH₄Cl:<br /><br />\[ \text{Mass of NH₄Cl} = \text{Moles of NH₄Cl} \times \text{Molar mass of NH₄Cl} \]<br />\[ \text{Mass of NH₄Cl} = 0.150 \, \text{mol} \times 53.50 \, \text{g/mol} = 8.025 \, \text{g} \]<br /><br />Therefore, the mass of ammonium chloride in the sample is \( 8.025 \, \text{g} \).