Question

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?

Answer 1

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

Step-by-step 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