3ZnBr2+2NoParrow Zn3P2+2NoBr3 How many grams of NoBr3 do I get from 164.2 grams of ZnBr2? (Zn=65Br=80No=259) 243 g 544 g 1088 g 365g

To solve this problem, we need to use the concept of stoichiometry and the molar mass of the compounds involved.<br /><br />Given information:<br />- The balanced chemical equation is: $3ZnBr2 + 2NoP \rightarrow Zn3P2 + 2NoBr3$<br />- The molar masses are: $Zn = 65$, $Br = 80$, $No = 259$<br /><br />Step 1: Calculate the molar mass of $ZnBr2$.<br />Molar mass of $ZnBr2 = 65 + 2(80) = 225$ g/mol<br /><br />Step 2: Calculate the number of moles of $ZnBr2$.<br />Number of moles of $ZnBr2 = \frac{164.2 \text{ g}}{225 \text{ g/mol}} = 0.728 \text{ mol}$<br /><br />Step 3: Use the stoichiometry of the balanced equation to find the number of moles of $NoBr3$.<br />According to the balanced equation, 3 moles of $ZnBr2$ produce 2 moles of $NoBr3$.<br />Number of moles of $NoBr3 = \frac{2}{3} \times 0.728 \text{ mol} = 0.485 \text{ mol}$<br /><br />Step 4: Calculate the molar mass of $NoBr3$.<br />Molar mass of $NoBr3 = 259 + 3(80) = 559$ g/mol<br /><br />Step 5: Calculate the mass of $NoBr3$ produced.<br />Mass of $NoBr3 = 0.485 \text{ mol} \times 559 \text{ g/mol} = 270.435 \text{ g}$<br /><br />Therefore, the correct answer is 270.435 g of $NoBr3$.<br /><br />The closest option to this value is 243 g, so the correct answer is 243 g.