2YbS+CfCl4-gt 2YbCl2+CfS2 How many grams of YbCl2 do I get from 96g YbS and 17 4gCfCl4(Yb=173S=32Cf=251Cl=35) 108 g 114g 216 g 227 g

To determine the amount of YbCl2 produced from 96g of YbS and 17.4g of CfCl4, we need to follow these steps:<br /><br />1. Calculate the molar masses of YbS, CfCl4, and YbCl2.<br />2. Convert the given masses of YbS and CfCl4 to moles.<br />3. Determine the limiting reactant.<br />4. Calculate the theoretical yield of YbCl2.<br /><br />Given:<br />- Yb = 173 g/mol<br />- S = 32 g/mol<br />- Cf = 251 g/mol<br />- Cl = 35.5 g/mol<br /><br />Molar masses:<br />- YbS = 173 + 32 = 205 g/mol<br />- CfCl4 = 251 + (4 × 35.5) = 351.5 g/mol<br />- YbCl2 = 173 + (2 × 35.5) = 244 g/mol<br /><br />Step 1: Convert the given masses to moles.<br />Moles of YbS = 96 g / 205 g/mol = 0.468 mol<br />Moles of CfCl4 = 17.4 g / 351.5 g/mol = 0.0495 mol<br /><br />Step 2: Determine the limiting reactant.<br />The balanced chemical equation is:<br />2 YbS + CfCl4 → 2 YbCl2 + CfS2<br /><br />From the balanced equation, we see that 2 moles of YbS react with 1 mole of CfCl4. Therefore, YbS is the limiting reactant.<br /><br />Step 3: Calculate the theoretical yield of YbCl2.<br />Since YbS is the limiting reactant, we can use its moles to calculate the yield of YbCl2.<br />Moles of YbCl2 produced = 2 × 0.468 mol = 0.936 mol<br /><br />Step 4: Convert the moles of YbCl2 to grams.<br />Mass of YbCl2 = 0.936 mol × 244 g/mol = 229.464 g<br /><br />Therefore, the amount of YbCl2 produced from 96g of YbS and 17.4g of CfCl4 is approximately 229.464 grams.