Question

Molten gallium reacts with arsenic to form the semiconductor, gallium arsenide, GaAs, used in light emitting diodes and solar cells: Ga(l) + As(s) → GaAs(s) If 4.00 g of gallium is reacted with 5.50 g of arsenic how many grams of the excess reactant are left at the end of the reaction?

Answer 1

Answer:

1.2 g

Explanation:

For gallium:-

Mass of gallium= 4.00 g

Molar mass gallium = 69.723 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{4.00\ g}{69.723\ g/mol}$

$Moles\ of\ gallium= 0.0574\ mol$

For arsenic:-

Mass of arsenic = 5.50 g

Molar mass of arsenic = 74.9216 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{5.50\ g}{74.9216\ g/mol}$

$Moles\ of\ arsenic= 0.0734\ mol$

According to the given reaction:

$Ga_{(l)}+As_{(s)}\rightarrow GaAs_{(s)}$

1 mole of gallium react with 1 mole of Arsenic

0.0574 mole of gallium react with 0.0574 mole of Arsenic

Moles of arsenic required = 0.0574 moles  

Available moles of arsenic = 0.0734 moles  

Limiting reagent is the one which is present in small amount. Thus, gallium is limiting reagent.

Moles left unreacted of arsenic = 0.0734 moles  - 0.0574 moles  = 0.016 moles

Mass = Moles * Molar mass = 0.016 * 74.9216 g = 1.2 g