Question 9 (1 point) Determine the theoretical yield of H2S (in moles) if 4.0 mo

Question

Question 9 (1 point) Determine the theoretical yield of H2S (in moles) if 4.0 mol Al2S3 and 4.0 m reacted according to the following balanced reaction. A possibly useful molan Al2S3 150.17 g/mol. ol H 20 are mass is Al53(+6 H22AOH)3(e)+3 H 58) sona, 5:53PM Qui CHMT020 INTRO TO CHEMISTRY ENHANCED 593939-DCOnline 12 mol H2S 4.0 mol H29 18 mol H2S 6.0 mol H2S 2.0 mol H2S Save Question 10 (1 point) Determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N204 and 45.0 g N2H4. Some possibly useful molar masses are as follows: N204 92.02 g/mol, N2H4 32.05 g/mol. N2040) +2N2H40)3N2)+4H28) LR N2H. 59.0 g N2 formed LR N204, 105 g N2 formed LR N204, 45.7 g N2 formed LR N2H4, 13.3 g N2 formed No LR, 45.0 g N2 formed Save

Explanation / Answer

9   Determine the number of moles of one of the products produced if all of each reactant is used up.

   = 4 mol Al2S3 x 3 mol H2S/1 mol Al2S3 = 12 mol H2S

   = 4 mol H20 x 3 mol H2S/6 mol H2O = 2 mol H2S

Use the smallest number of moles of the product (H2S) to calculate the theoretical yield of product (H2S).

Theoretical Yield = 2 mol H2S

Note: Since the reactant, H2O, produces the least amount of product, it is the limiting reactant and the other    reactant, Al2S3, is in excess.

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