Question

Consider the following unbalanced equation. How many Liters of bromine are needed to produce 12 moles of Aluminum bromide? The density of bromine is 3.1 g/mL.

Al (s) + Br2 (l)= AlBr3 (s)

Answer 1

`V=929mL`

Step-by-step explanation:

Hello,

In this case, the balanced chemical reaction is:

`2Al (s) + 3Br_2 (l)\rightarrow 2AlBr_3 (s)`

In such a way, we use the 2:3 molar ratio between aluminum bromide and bromine and its atomic mass which is 160 g/mol to find the grams of bromine that are produced:

`m_(Br_2)=12molAlBr_3*(3molBr_2)/(2molAlBr_3) *(160gBr_2)/(1molBr_2) =2880gBr_2`

Then we compute the volume:

`V=(m)/(\rho)=(2880gBr_2)/(3.1g/mL)\\ \\V=929mL`

Regards.

Answer 2

0.92787 liters of bromine are needed to produce 12 moles of aluminum bromide.

Step-by-step explanation:

You have the following balanced equation:

2 Al (s) + 3 Br₂ (l) ⇒ 2 AlBr₃ (s)

First of all, the following rule of three should be applied to know the amount of moles of bromine needed: if 2 moles of aluminum bromide are produced by stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction) for 3 mole of bromine, 12 moles of aluminum bromide with how many moles of bromine are produced?

`moles of bromine=(12 moles of aluminum bromide*3 mole of bromine)/(2 mole of aluminum bromide)`

moles of bromine= 18

Being the molar mass of the bromine Br₂ 159.8 g/mol then the mass of 18 moles of Br₂ is:

18 moles* 159.8 g/mol= 2,876.4 grams

Density is a property that indicates the amount of mass per unit volume. Then the following rule of three applies: if by the definition of density 3.1 grams of bromine are present in 1 mL, 2,876.4 grams of bromine are present in how much volume is it?

`volume=(2,876.4 grams*1mL)/(3.1 grams)`

volume= 927.87 mL

Being 1,000 mL= 1 L, then 927.87 mL= 0.92787 L

0.92787 liters of bromine are needed to produce 12 moles of aluminum bromide.