22 Cl_(2)+2KBrarrow Br_(2)+2KCl How many grams of potassium chloride (KCl) can be produced from 356 g of potassium bromide (KE 7499 2219 479 g 814 g

To determine how many grams of potassium chloride (KCl) can be produced from 356 g of potassium bromide (KBr), we need to follow these steps:<br /><br />1. **Write down the balanced chemical equation:**<br /> \[<br /> Cl_{2} + 2KBr \rightarrow Br_{2} + 2KCl<br /> \]<br /><br />2. **Calculate the molar masses of the reactants and products:**<br /> - Molar mass of KBr: \(39.10 \, \text{g/mol (K)} + 79.90 \, \text{g/mol (Br)} = 119.00 \, \text{g/mol}\)<br /> - Molar mass of KCl: \(39.10 \, \text{g/mol (K)} + 35.45 \, \text{g/mol (Cl)} = 74.55 \, \text{g/mol}\)<br /><br />3. **Convert the mass of KBr to moles:**<br /> \[<br /> \text{Moles of KBr} = \frac{356 \, \text{g}}{119.00 \, \text{g/mol}} = 2.983 \, \text{mol}<br /> \]<br /><br />4. **Use the stoichiometry of the balanced equation to find the moles of KCl:**<br /> According to the balanced equation, 2 moles of KBr produce 2 moles of KCl. Therefore, the moles of KCl produced will be the same as the moles of KBr:<br /> \[<br /> \text{Moles of KCl} = 2.983 \, \text{mol}<br /> \]<br /><br />5. **Convert the moles of KCl to grams:**<br /> \[<br /> \text{Mass of KCl} = 2.983 \, \text{mol} \times 74.55 \, \text{g/mol} = 222.5 \, \text{g}<br /> \]<br /><br />Therefore, the correct answer is:<br />\[<br />\boxed{221.9 \, \text{g}}<br />\]