QUESTION 1 with 300. mL of o.100M Ca(OH)2 which undergo the following reaction 0

Question

QUESTION 1 with 300. mL of o.100M Ca(OH)2 which undergo the following reaction 00MlH2 All problems in this homework refer to mixing 100. ml of 0.200 M AlIS04)3 Al2(SO4)3 (aq) + 3 Ca(OH)2 (aq) 2 Al(OH)3 (s) + 3 CaS04 (s) Which is the limiting reactant? OAl2(S04)3 OCa(OH)2 AIOH)3 O CaS04 QUESTION 2 All problems in this homework refer to mixing 100. ml of 0.200 M Ai2 (S04)3 Al2(SO4)3 (aq) + 3 Ca(OH2 (aq)- 2 Al(OH)3 (s) + 3 CaSO4 (s) How many moles of Al2(SO4)3 remain? with 300. mL of 0.100MCa(OH)2 which undergo the following reaction QUESTION 3 All problems in this homework refer to mixing 100. mL of 0 200 M Al2(So4)3 with 300. mL of o.100M Ca(OH)2 which undergo the following reaction: Al2(SO4)3 (aq) + 3 Ca(OH)2 (aq)-> 2 AKOH)3 (s) + 3 CaSO4 (s) How many moles of Ca(OH)2 ill remain?

Explanation / Answer

Al2(SO4)3(aq) + 3Ca(OH)2(aq) -----------> 2Al(OH)3 (s) + 3CaSO4(s)
no of moles of Al2(SO4)3 = molarity * volume in L
                          = 0.2*0.1 = 0.02 moles
no of moles of Ca(OH)2   = molarity * volume in L
                         = 0.1*0.3 = 0.03 moles
1 moles of Al2(SO4)3 react with 3 moles of Ca(OH)2
0.02 moles of Al2(SO4)3 react with = 3*0.02/1 = 0.06 moles of Ca(OH)2
    Ca(OH)2 is limiting reactant
Ca(OH)2>>>>>answer
3 moles of Ca(OH)2 react with 1 moles of Al2(SO4)3
0.03 moles of Ca(OH)2 react with = 1*0.03/3 = 0.01 moles of Al2(SO4)3

Q2
Al2(SO4)3 is excess reactant
no moles of Al2(SO4)3 remains = 0.02-0.01 = 0.01 moles

Q3
no of moles of ca(OH)2 remain   = 0

Q4
3 moles of ca(OH2 react with Al2(SO4)3 to gives 2 moles of Al(OH)3
0.03 moles of Ca(OH)2 react with Al2(SO4)3 to gives= 2*0.03/3 = 0.02 moles of Al2(SO4)3

Q5
percentage yield = actual yield*100/theoritical yield
                  = 0.015*100/0.02 = 75% >>>>answer

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.