2AgNO_(3)+BaCl_(2)arrow 2AgCl+Ba(NO_(3))_(2) How many grams of silver chloride are produced from 5.0 g of silver nitrate reacting with an excess of barium chloride?

To determine how many grams of silver chloride (AgCl) are produced from 5.0 g of silver nitrate (AgNO₃) reacting with an excess of barium chloride (BaCl₂), we need to follow these steps:<br /><br />1. **Write the balanced chemical equation:**<br /> \[<br /> 2AgNO_3 + BaCl_2 \rightarrow 2AgCl + Ba(NO_3)_2<br /> \]<br /><br />2. **Calculate the molar masses of the reactants and products:**<br /> - Molar mass of AgNO₃: \( 107.87 \, (\text{Ag}) + 14.01 \, (\text{N}) + 3 \times 16.00 \, (\text{O}) = 169.87 \, \text{g/mol} \)<br /> - Molar mass of AgCl: \( 107.87 \, (\text{Ag}) + 35.45 \, (\text{Cl}) = 143.32 \, \text{g/mol} \)<br /><br />3. **Convert the mass of AgNO₃ to moles:**<br /> \[<br /> \text{Moles of AgNO}_3 = \frac{\text{Mass of AgNO}_3}{\text{Molar mass of AgNO}_3} = \frac{5.0 \, \text{g}}{169.87 \, \text{g/mol}} \approx 0.0294 \, \text{mol}<br /> \]<br /><br />4. **Use the stoichiometry of the balanced equation to find the moles of AgCl produced:**<br /> According to the balanced equation, 2 moles of AgNO₃ produce 2 moles of AgCl. Therefore, the moles of AgCl produced will be the same as the moles of AgNO₃ used:<br /> \[<br /> \text{Moles of AgCl} = 0.0294 \, \text{mol}<br /> \]<br /><br />5. **Convert the moles of AgCl to grams:**<br /> \[<br /> \text{Mass of AgCl} = \text{Moles of AgCl} \times \text{Molar mass of AgCl} = 0.0294 \, \text{mol} \times 143.32 \, \text{g/mol} \approx 4.21 \, \text{g}<br /> \]<br /><br />Therefore, 4.21 grams of silver chloride (AgCl) are produced from 5.0 grams of silver nitrate (AgNO₃) reacting with an excess of barium chloride (BaCl₂).