1. 15.6 g of aluminum sulfide reacts with 10.0 g of water according to the reaction below: _Al₂S, (s) + H₂O (1)→ Al(OH), (aq) +_____ H₂S (g) a. Which substance is the limiting reactant? b. How many liters of H₂S is formed? c. What amount of excess reactant is left over, in grams?

Introductory Chemistry: A Foundation
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Chapter9: Chemical Quantities
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How to determine LK:
1. Convert reactant quantities to mol
2. Divide each mol amount by balanced coefficient
Avogadro's # = 6.023 * 10²³ molecules/1 mol
STP Volume = 1 mol/22.4 L
3. Smallest value = LR
Left over = initial - consumed
1. 15.6 g of aluminum sulfide reacts with 10.0 g of water according to the reaction below:
_Al₂S₂ (s) + _H₂O (1)→ ___Al(OH)₂ (aq) +
—H₂S (8)
a. Which substance is the limiting reactant?
b. How many liters of H₂S is formed?
c. What amount of excess reactant is left over, in grams?
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Transcribed Image Text:How to determine LK: 1. Convert reactant quantities to mol 2. Divide each mol amount by balanced coefficient Avogadro's # = 6.023 * 10²³ molecules/1 mol STP Volume = 1 mol/22.4 L 3. Smallest value = LR Left over = initial - consumed 1. 15.6 g of aluminum sulfide reacts with 10.0 g of water according to the reaction below: _Al₂S₂ (s) + _H₂O (1)→ ___Al(OH)₂ (aq) + —H₂S (8) a. Which substance is the limiting reactant? b. How many liters of H₂S is formed? c. What amount of excess reactant is left over, in grams? Page 1 / 2 +
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