Question

For the following reaction, 65.4 grams of barium hydroxide are allowed to react with 35.7 grams of sulfuric acid. barium hydroxide (aq) + sulfuric acid (aq)»barium sulfate (s) + water (0) hat is the maximum amount of barium sulfate that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete?grams Submit Answer For the following reaction, 22.6 grams of phosphorus (P) are allowed to react with 82.1 grams of chlorine gas. phosphorus (P) (s)+ chlorine (g)»phosphorus trichloride (I) What is the maximum amount of phosphorus trichloride that can be formed? What is the FORMULA for the limiting reagent? grams #1 What amount of the excess reagent remains after the reaction is complete? grams Submit Answer

Answer 1

Balanced equation:
Ba(OH)₂ + H₂SO₄ ====> BaSO₄ + 2 H₂O
Reaction type: double replacement

Moles of Ba(OH)2 = 65.4 / 171.34 = 0.3816 Moles

Moles of H2SO4 = 35.7 / 98.07 =  0.36399 Moles

Limiting reagent is H2SO4

Mass of BaSO4 produced =  0.36399 x 233.38 = 84.95 gm

Excess amount of Ba(OH)2 = 65.4 - 62.367 = 3.033 gm

Question 2

Balanced equation:
P₄ + 6 Cl₂ ====> 4 PCl₃
Reaction type: synthesis

22.6 gm of P4 = 22.6 / 123.89 =  0.1824 Moles

82.1 gm of Cl2 = 82.1 / 70.906 =  1.157 Moles

Limiting reagent is P4

Mass of PCl3 formed = 0.7296 x 137.33 = 100.20 gm

Excess Cl2 present after reaction = 82.1 - 77.60 = 4.5 gm