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If he has an excess of ammonium sulfide and 125.0 mL of a 1.51 M solution of Al(NO3)3, how many
grams of aluminum sulfide can he make?

Sagot :

Eduard22sly

The mass of the aluminum sulfide, Al₂S₃, that can be produced from the reaction is 14.16 g

How to determine the mole of Al(NO₃)₃

  • Volume = 125 mL = 125 / 1000 = 0.125 L
  • Molarity of Al(NO₃)₃ = 1.51 M
  • Mole of Al(NO₃)₃ =?

Mole = Molarity x Volume

Mole of Al(NO₃)₃ = 1.51 × 0.125

Mole of Al(NO₃)₃ = 0.18875 mole

How to determine the mole of Al₂S₃ produced

Balanced equation

2Al(NO₃)₃ + 3(NH₄)₂S → Al₂S₃ + 6NH₄NO₃

From the balanced equation above,

2 moles of Al(NO₃)₃ reacted to produce 1 mole of Al₂S₃

Therefore,

0.18875 mole of Al(NO₃)₃ will react to produce = 0.18875 / 2 = 0.094375 mole of Al₂S₃

How to determine the mass of Al₂S₃ produced

  • Mole of Al₂S₃ = 0.094375 mole
  • Molar mass of Al₂S₃ = (27×2) +(32×3) = 150 g/mol
  • Mass of Al₂S₃ =?

Mass = mole × molar mass

Mass of Al₂S₃ = 0.094375 × 150

Mass of Al₂S₃ = 14.16 g

Learn more about stoichiometry:

https://brainly.com/question/14735801

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