PoS3+6FrBrarrow PoBr6+3Fr2S How many grams of Fr2S do I get from 252.3 grams of PoS3? (Po=209S=32Fr=223) 478 g 395 g 1434 g 1185 g

To solve this problem, we need to use the concept of stoichiometry and the given molar masses of the compounds involved.<br /><br />Given information:<br />- The balanced chemical equation is: $PoS3 + 6FrBr \rightarrow PoBr6 + 3Fr2S$<br />- The molar masses are: $Po = 209$, $S = 32$, $Fr = 223$<br /><br />Step 1: Calculate the molar mass of $PoS3$.<br />Molar mass of $PoS3 = 209 + (3 × 32) = 359$ g/mol<br /><br />Step 2: Calculate the molar mass of $Fr2S$.<br />Molar mass of $Fr2S = (2 × 223) + 32 = 478$ g/mol<br /><br />Step 3: Calculate the number of moles of $PoS3$.<br />Number of moles of $PoS3 = \frac{252.3 \text{ g}}{359 \text{ g/mol}} = 0.703 \text{ mol}$<br /><br />Step 4: Use the stoichiometry of the balanced equation to find the number of moles of $Fr2S$.<br />According to the balanced equation, 1 mole of $PoS3$ produces 3 moles of $Fr2S$.<br />Number of moles of $Fr2S = 0.703 \text{ mol} × 3 = 2.109 \text{ mol}$<br /><br />Step 5: Calculate the mass of $Fr2S$.<br />Mass of $Fr2S = 2.109 \text{ mol} × 478 \text{ g/mol} = 1008.732 \text{ g}$<br /><br />Therefore, the correct answer is 1008.732 g.