Question

how many moles of lithium nitrate are theoretically produced if we start with 34 moles of Ca(NO3)2 and 24 moles of LigPO4? Reaction: 3Ca(NO3)2 + 2Li3PO4 -- LINO3 + Caz(PO4)2 A) 1.1 B) 6.8 C) 1.2 D) 7.2 E) not enough information 28) – 28) What is the limiting reactant for the following reaction given we have 2.6 moles of HCl and 1.4 moles of Ca(OH)2? Reaction: 2HCl + Ca(OH)2 + 2H2O +CaCl2 A) Ca(OH)2 B) HCI C) CaCl2 D) H0 E) not enough information 29) 29) What is the limiting reactant for the reaction below given that you start with 10.0 grams of Al and 19.0 grams of O2? Reaction: 4A1 +30 2-2Al2O3 A) 02 B) AI C) Al2O3 D) both 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 +502 -- 4 NO+6 H20 A) 70% B) 60% C) 15% D) 30% E) none of the above 311 31) Consider the following reaction: 2 Mg +022 MgO AH rxn--1203 kJ Calculate the amount of heat (in kJ) associated with complete reaction of 4 moles of Mg A) -1203 kJ B) -4812 kJ C) -2406 kJ D) -601.5 kJ 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

28)

2 mol of HCl reacts with 1 mol of Ca(OH)2

for 2.6 mol of HCl, 2.6/2 = 1.3 mol of Ca(OH)2 is required

We have more Ca(OH)2 than that.

So, HCl is limiting reagent

Answer: B

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