When benzene (c6h6) reacts with bromine (br2) bromobenzene(c6h5br) is obtained: c6h6 + br2 → c6h5br + hbr what is the theoretical yield of bromobenzene in this reaction when 30.0g of benzene reacts with 65.0 g of bromine? b. if the actual yield of bromobenzene was 56.7 g what was the percentage yield?

Answer A) : We have to calculate the number of moles of Benzene involved in the reaction,

30 g / 78 moles of benzene = 0.384 moles

For bromine it will be the same process,

65 g / 159.8 moles = 0.406 moles

By observing the reaction given above we can say that the reaction ratio of bromine and benzene is 1 : 1

We need to find the mass of bromobenzene,

which should be, 6(12) + 5 (1)+ 79.90 = 156.9 g/mol

So, the mass of bromobenzene will be 156.9 g/mol X 0.3846 mol = 60.343 g

Hence the theoretical yield will be 60.34 g

Answer B) : To calculate the actual yield we have to divide it with theoretical yield.

(56.7g / 60.343 g ) X100% = 93.96 %

Here, we can say that we got 93.96 % of actual yield.

As we know it is impossible to get 100% yield in any reaction.

answer

i think the answer is a) as air pressure in an area increases, the density of the gas particles in that area decreases.

hope it

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.