Question

a)

According to the following reaction, how many moles of bromine trifluoride are necessary to form 0.841 moles fluorine gas?

bromine trifluoride (g) →bromine(g) + fluorine(g)

_______moles bromine trifluoride

b) According to the following reaction, how many moles of dichloromethane (CH₂Cl₂) will be formed upon the complete reaction of 0.227 moles methane (CH₄) with excess carbon tetrachloride?

methane (CH₄) (g) +carbon tetrachloride (g)→dichloromethane (CH₂Cl₂) (g)

________moles dichloromethane (CH₂Cl₂)

Answer 1

(a) from the balanced equation we see that

2 moles of BrF3 gives 3 moles of Fluorine gas

So 0.841 moles of Fluorine gas will be given by 2/3* 0.841 moles of BrF3 = 0.56 moles of boron trifluoride.

(b) 1 mole of methane forms 2 moles of dichloromethane

So 0.227 moles of CH4 will form 2/1*0.227= 0.454 moles of.dichloromethane.