Question

Reaction of aluminum sulfide with water provides aluminum hydroxide and H2S(g). If 20.00 g of aluminum sulfide is combined with 2.00 mL of water (d = 1.000 g/mL), what amount remains of the reagent present in excess? Report your answer to the appropriate number of significant figures. g

Answer 1

Answer:

Explanation:

A mole ratio is ​the ratio between the amounts in moles of any two compounds involved in a chemical reaction. The mole ratio can be determined by examining the coefficients in front of formulas in a balanced chemical equation.

Step 1: write the balanced chemical equation.

Al₂S₃(s) + 6H₂O(l) -----> 2Al(OH)₃(aq) + 3H₂S(g)

Step 2: calculate the moles of Al₂S₃(s) and  H₂O(l)

Moles of Al₂S₃= mass given / molar mass = ( 20 g / 150.158 g/mol ) = 0.1332 mol

Moles of H₂O = ( 2 g / 18.01528 g/mol ) = 0.111 mol [ mass =density × volume = (1 g/mL × 2 mL ) = 2 g ]

Step 3: Determine the limiting reagent

Al₂S₃(s) + 6H₂O(l) -----> 2Al(OH)₃(aq) + 3H₂S(g)

According to the reaction we need
6 mol of H₂O requires for 1 mol of Al₂S₃
so, for 0.111 mol of H₂O we will need =(1 mol of Al₂S₃ / 6 mol of H₂O )× 0.111 mol of H₂O = 0.0185 mol of Al₂S₃
Since we need only 0.0185 mol of Al₂S₃ hence Al₂S₃ is in excess ( since given = 0.1332 mol ) and H₂O limiting reagent.

so the amount of product will formed according to the limiting reagent ( i.e H₂O amount )

(c) Step 4: Calculation of excess reagent   Al₂S₃ left:

we know moles of Al₂S₃ used = 0.1332  mol - 0.0185  mol = 0.1147  mol

hence, mass remaining after reaction in excess = 0.1147  mol × 150.158 g/mol = 17.2 grams