MolSynapse
question bank
All chapters
NEET
JEE
Practice
MolSynapse
/
Question bank
/
Mole Concept
/
Limiting reagent, percent yield and percent excess
NEET level
Single correct option
Difficulty 1 of 3
mc186
50
g
of
CaCO
X
3
is heated:
CaCO
X
3
CaO
+
CO
X
2
. If
21
g
of
CaO
is actually obtained, the percentage yield is (Ca = 40, C = 12, O = 16)
A
75%
B
42%
C
60%
D
84%
Answer: A
n(CaCO3) = 0.50 mol, so theoretical CaO = 0.50 x 56 = 28 g. Percentage yield = 21/28 x 100 = 75%.
Practice Mole Concept
More on Limiting reagent, percent yield and percent excess
Previous question
Next question
More questions on Limiting reagent, percent yield and percent excess
17 g of NH3 and 48 g of O2 react as far as possible according to 4NH3 + 5O2 -> 4NO + 6H2O (N=14, O=16, H=1). Choose the correct statement(s)...
JEE Advanced level
28 g of N2 reacts with 10 g of H2 according to N2 + 3H2 -> 2NH3. The maximum mass of NH3 formed is (N = 14, H = 1)
NEET level
In 2Mg + O2 -> 2MgO, 4.8 g of Mg is heated with 1.6 g of O2. The mass of MgO formed is (Mg = 24, O = 16)
NEET level
4 mol of H2 is mixed with 3 mol of O2 for 2H2 + O2 -> 2H2O. The percentage of O2 in excess over the amount required is
NEET level
Fe2O3 + 3CO -> 2Fe + 3CO2. When 240 g of Fe2O3 reacts with 84 g of CO, the maximum mass of Fe obtained is (Fe = 56, C = 12, O = 16)
JEE Main level
5.4 g of Al reacts with 18.25 g of HCl: 2Al + 6HCl -> 2AlCl3 + 3H2. The volume of H2 at STP (22.4 L/mol) is (Al = 27, H = 1, Cl = 35.5)
JEE Main level