A HACC student started with 0.8656 g of copper (II) chloride CuCb. After reactin

Question

A HACC student started with 0.8656 g of copper (II) chloride CuCb. After reacting it with water vapor, she weighed 1.0255 g of her product, copper (II) chloride dihydrate CuCh 2H2O. What was her percent yield? 4. The formula of the product uses the notation that is used for "hydrates", which are solids that incorporate a specific number of water molecules into their crystal structures. A "dihydrate" has exactly 2 water molecules for each CuCl2 molecule. We encountered this notation in lab- and you should be familiar with it.) CuCl2 + 2 H2O CuCh 2H20

Explanation / Answer

The number of moles of copper chloride used = 0.8656 / molar mass

= 0.8656 / 134.45

= 0.0064 moles

So according to the well balanced reaction , the number of moles hydrate formed will also be 0.0064 moles

Theoretical yield = 0.0064 x molar mass of hydrate

= 0.0064 x 170.48

= 1.091 g

Percent yield = ( actual yield / theoretical yield) x 100

= ( 1.0255 / 1.091 ) x 100

= 93.99 %

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.