Molten gallium reacts with arsenic to form the semiconductor, gallium arsenide, GaAs, used in light-emitting diodes and solar cells: Ga (I) + As (s) GaAs (s)

a. 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?

b. If 4.00 g of gallium is reacted with 4.94 g of arsenic, how many grams of the excess reactant are left at the end of the reaction?

c. If 4.00 g of gallium is reacted with 0.56 g of arsenic, how many grams of the excess reactant are left at the end of the reaction?

d. If 8.94 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?

e. If 4.00 g of gallium is reacted with 1.50 g of arsenic, how many grams of the excess reactant are left at the end of the reaction?

a) 1.2g of arsenic b) 0.64g of arsenic c) 3.481g of gallium d) 3.806g of gallium e) 2.61g arsenic

Explanation:

The balanced equation is:

Ga + As = GaAs, 1:1 mole ratio

a) mass (gallium) / molar mass of Ga = 4 / 69.723 = 0.0574mol

Mass (arsenic) / molar mass of As = 5.5/74.9216 = 0.0734, subtracting the moles from each other (knowing already that the ratio is 1:1), arsenic is in excess by 1.2g

b) repeating the procedure (changing the values)

It will be 0.0574 to 0.06593

Arsenic is in excess by 0.00854

0.00854 * mass of arsenic (such must be done for the first remaining mole) = 0.64g

c) the mole ratio is 0.0574: 0.00747

Gallium is in excess by 0.05

Mass of excess gallium = 0.05 * 69.723 = 3.481g

d) using the mass given, the new ratio is

0.128: 0.0734

Gallium is in excess by 0.054mol

Mass of excess gallium = 0.054*69.723 = 3.806g

e) using the mass again, the new ratio is 0.0574: 0.02,

Gallium is in excess by 0.0374*69.723 = 2.61g