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

Answer:

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

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?

$moles of bromine=\frac{12 moles of aluminum bromide*3 mole of bromine}{2 mole of aluminum bromide}$

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?

$volume=\frac{2,876.4 grams*1mL}{3.1 grams}$

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.

Answer 2

Answer:

$V=929mL$

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:

$m_{Br_2}=12molAlBr_3*\frac{3molBr_2}{2molAlBr_3} *\frac{160gBr_2}{1molBr_2} =2880gBr_2$

Then we compute the volume:

$V=\frac{m}{\rho}=\frac{2880gBr_2}{3.1g/mL}\\ \\V=929mL$

Regards.