Question

Calculate the total masses of the reactants for the following equation:2Al(s)+3Br2(g)→2AlBr3(g)

Answer 1

To find the mass of the reactants we must convert the moles to grams, to do this conversion we use the molar mass of the compounds.

Molar mass Al=26.98g/mol

Molar mass Br2= 79.90x2 = 159.80g/mol

Now, assuming that the moles that we have are the moles that we have due to the stoichiometry of the reaction, we will have 2 moles of Aluminum and 3 moles of Bromine. So the mass of the compound will be:

Mass of Al

`\begin{gathered} gAl=givenmolAl*(MolarMass,gAl)/(1molAl) \\ gAl=2molAl*(26.98gAl)/(1molAl)=53.96gAl \end{gathered}`

Mass Br2

`\begin{gathered} gBr_2=GivenmolBr_2*(MolarMass,gBr_2)/(1molBr_2) \\ gBr_2=3molBr_2*(159.80gBr_2)/(1molBr_2)=479.4gBr_2 \end{gathered}`

Now, the total mass will be equal to the sum of the masses of the reactants. We have that:

`\begin{gathered} TotalMassesReactants=MassAl+MassBr_2 \\ TotalMassesReactants=53.96g+479.4g=533.36g \end{gathered}`

Answer: The total masses of the reactants will be 533.36g