Question

help with all qeustions(:

Prelab Ouestions: Name 1) What is the molar mass of KAI(SO.), -12H,O ? (the 12 waters are part of the molecule) 2) Calculate the mass of alum that can be produced from 14.471g of Al 3) Calculate the mass of alum that can be formed from 0.724 moles of KOH 4) Calculate the mass of alum that can be formed from 1.10 mole of H2SO4 5) a) Based on the data above which reactant is the limiting reactant? b) What is the theoretical yield of alum from using all of this limiting reactant? c) What is the percent yield if the actual yield of alum is 236.435 g

Answer 1

1. KAl(SO4)2 12H2O

Molar Mass of K = 39amu

Molar Mass of Al = 27 amu

Molar Mass of S = 32 amu

Molar Mass of O = 16 amu

Molar Mass of H = 1 amu

Total Molar Mass = 39 + 27 +2(32) + 8(16) +12(18)

= 474 amu

2. General equation of formation of Alum is-

Al + KOH + 2 H2SO4 + H2O $\rightarrow$ Alum

Thus, 1 mole Al : 1 mole Alum

No of moles = Given wt / Mol wt

Moles of Al = 14.471/ 27 = 0.535 moles

As a result 0.535 moles of Alum will be produced

0.535 = Given Wt / 474

Wt = 254 g

Thus, 254 g of Alum will be produced from 14.471g Al.

3. 1 mole KOH : 1 mole Alum

0.724 moles KOH : 0.724 moles Alum

0.724 = Given Wt/474

Wt = 343.176

Thus, 0.724 moles of KOH will produce 343.176 g Alum.

4. 2 moles H2SO4 : 1mole Alum

1.10 mole : 0.55 mole Alum

0.55 = Given Wt/ 474

Wt = 260.7 g

Thus, 260.7g Alum will be peoduced from 1.10 moles of acid.