Answer to Question #299124 in General Chemistry for Sav

Question

A container of 500.00 g of Br 2 was reacted with 87.93 g of carbon. What is the limiting reactant? What mass of product was expected to form? If 352.77 g of CBr4 actually formed, what is the percent yield?

C +    Br2 —>     CBr4

Answer 1

• First. we balance the equation: C +    2Br₂ —>     CBr₄
• In order to find the limiting reactant, we must find the moles of each.
• Moles are given by mass divided by molar mass
• For Br₂ we get 500/160 which is 3.125 moles
• For C we get 87.93/12 which is 7.3275 moles
• Comparing the two, It is clear that carbon is in excess and therefore, Br₂ is the limiting reactant.
• In order to get the mass of product, we use the moles of Br₂ and mole ratio such that the moles of CBr₄ are half those of Br₂. This will be 3.125/2 which gives 1.5625moles.
• Convert moles to mass: 332g=1mole what about 1.5625moles.
• 1.5625*332 which gives 518.75g.
• To find the percentage yield, take the actual yield divided by the theoretical value and multiply by 100%.
• 352.77/518.78*100% which gives 67.9999%