Question

27) How many moles of lithium nitrate are theoretically produced if we start with 34 moles Ca(NO₃)₂ and 24 moles of Li₃PO₄ ?

A) 1.4

B) 6.8

C) 12

D) 7.2

E) not enough information

28) What is the limiting reactant for the following reaction given we have 2.6 moles of HCI and 1.4 moles of Ca(OH)₂?

Reaction: 2 HCl+Ca(OH₂ → 2 H₂ O+CaCl₂.

A) Ca(OH)₂

B) HCl

C) CaCl₂

D) H₂ O

E) not enough information

29) What is the limiting reactant for the reaction below given that you start with 10,0 grams of AI and 19.0 grams of O₂ ?

Reaction: 4 Al+3 O₂ → 2 Al₂ O₃

A) O₂

B) Al

C) Al₂ O₃

D) bath Al and O₂

E) not enough information

30) A sample of 8.5 g NH₃ on oxidation produces 4.5 g of NO. Calculate the percent yield. Reaction: 4 NH₃+5 O₂ → 4 NO+6 H₂ O

A) 70 %

B) 60 %

C) 15 %

D) 30 %

E) none of the above

31) Consider the following reaction: .2 Mg+O₂ → 2 MgO Δ HrOn=-1203 k]

Calculate the amount of heat (in kJ ) associated with complete reaction of 4 moles of Mg.

A) =1203 kg

B) -4512 kJ

C) -2406 kj

D) -601.5 k]

E) none of the above

Answer 1

27)

Balanced chemical equation is:

3 Ca(NO3)2 + 2 Li3PO4 ---> 6 LiNO3 + Ca3(PO4)2

3 mol of Ca(NO3)2 reacts with 2 mol of Li3PO4

for 3.4 mol of Ca(NO3)2, 2.267 mol of Li3PO4 is required

But we have 2.4 mol of Li3PO4

so, Ca(NO3)2 is limiting reagent

we will use Ca(NO3)2 in further calculation

According to balanced equation

mol of LiNO3 formed = (6/3)* moles of Ca(NO3)2

= (6/3)*3.4

= 6.8 mol

Answer: B