Question

Using the balanced equation `3Ba + Al2(SO_4)_3 -> 2Al + 3BaSO_4`, if 4g Ba react with 0.45 g `Al_2(SO_4)_3`, what is the excess reactant?

Answer 1

Aluminum sulfate produces the least amount of barium sulfate, so it is the limiting reactant and barium is the excess reactant.

Explanation:

Balanced Equation

`"3Ba + Al"_2"(SO"_4)_3"` `rarr` `"2Al + 3BaSO"_4`

The coefficients are the moles of each substance, and they can be used as molar ratios between two of the substances. No coefficient is understood to be `1`.

The reactant that produces the least amount of `"BaSO"_4"` is the limiting reactant.

You need the molar masses of each substance involved.

`"Ba":` `"137.327 g/mol"` (periodic table)
`"Al"_2"(SO"_4)_3":` `"342.131 g/mol"`
`"BaSO"_4:` `"233.383 g/mol"`
https://pubchem.ncbi.nlm.nih.gov/compound/24850
https://www.ncbi.nlm.nih.gov/pccompound/?term=%22BARIUM+SULFATE%22

The process for each element involves the following steps.

`color(red)("given mass Ba"` `rarr` `color(blue)("mol Ba"` `rarr` `color(green)("mol BaSO"_4"` `rarr` `color(purple)("mass BaSO"_4"`

`color(red)("given mass Al"_2("SO"_4")` `rarr` `color(blue)("mol Al"_2("SO"_4")"` `rarr` `color(green)("mol BaSO"_4"` `rarr` #color(purple) ("mass BaSO"_4"#

To go from given mass of reactant to moles reactant, divide the given mass by its molar mass by multiplying by the inverse of the molar mass. To go from mol reactant to mol product, multiply by the mol ratio that cancels the mol reactant. To go from mol product to mass product, multiply by its molar mass.

Reactant Barium

`color(red)(4"g Ba")xxcolor(blue)(1"mol Ba")/color(blue)(137.327"g Ba")xx(color(green)(3)color(green)("mol BaSO"_4))/color(green)(3"mol Ba")xxcolor(purple)(233.383"g BaSO"_4)/color(purple)(1"mol BaSO"_4)=color(purple)("7 g BaSO"_4)` (rounded to one sig fig due to 4 g)

Reactant `"Al"_2("SO"_4)_3`

`color(red)(0.45"g Al"_2("SO"_4)_3)xxcolor(blue)(1"mol Al"_2("SO"_4))/color(blue)(342.131"g Al"_2("SO"_4))xxcolor(green)(3"mol BaSO"_4)/color(green)(1"mol Al"_2("SO"_4))xxcolor(purple)(233.383"g BaSO"_4)/color(purple)(1"mol BaSO"_4)=color(purple)"0.92 g BaSO"_4`

Aluminum sulfate produces the least amount of barium sulfate, so it is the limiting reactant and barium is the excess reactant.