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28 January, 15:17

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?

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  1. 28 January, 15:43
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    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
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