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 = 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 = mole × molar mass
Mass of Al₂S₃ = 0.094375 × 150
Mass of Al₂S₃ = 14.16 g
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