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:1. Calculate the molar masses of YbS, CfCl4, and YbCl2.2. Convert the given masses of YbS and CfCl4 to moles.3. Determine the limiting reactant.4. Calculate the theoretical yield of YbCl2.Given:- Yb = 173 g/mol- S = 32 g/mol- Cf = 251 g/mol- Cl = 35.5 g/molMolar masses:- YbS = 173 + 32 = 205 g/mol- CfCl4 = 251 + (4 × 35.5) = 351.5 g/mol- YbCl2 = 173 + (2 × 35.5) = 244 g/molStep 1: Convert the given masses to moles.Moles of YbS = 96 g / 205 g/mol = 0.468 molMoles of CfCl4 = 17.4 g / 351.5 g/mol = 0.0495 molStep 2: Determine the limiting reactant.The balanced chemical equation is:2 YbS + CfCl4 → 2 YbCl2 + CfS2From the balanced equation, we see that 2 moles of YbS react with 1 mole of CfCl4. Therefore, YbS is the limiting reactant.Step 3: Calculate the theoretical yield of YbCl2.Since YbS is the limiting reactant, we can use its moles to calculate the yield of YbCl2.Moles of YbCl2 produced = 2 × 0.468 mol = 0.936 molStep 4: Convert the moles of YbCl2 to grams.Mass of YbCl2 = 0.936 mol × 244 g/mol = 229.464 gTherefore, the amount of YbCl2 produced from 96g of YbS and 17.4g of CfCl4 is approximately 229.464 grams.