Chemistry - Stoichiometry Question with Solution | TestHub
ChemistryStoichiometryCalculation of Percentage yieldMedium2 minQB
ChemistryMediummatching list
Match the list:
List - I | List - II |
|---|---|
(P) | (1) 300/7% carbon by mass |
(Q) CO | (2) Hydrocarbon |
(R) | (3) Vapour density = 40 |
(S) | (4) electrons in a mole ( Avogadro number) |
| (5) 50 % carbon atoms, by mole |
The CORRECT option is:
Options:
Answer:
B
Solution:
P) SO₃: Molar mass = g/mol. Vapour density (VD) = Molar mass / 2 = . So, P 3.
Q) CO: Molar mass = g/mol. % C by mass = %. So, Q 1.
R) C₆H₆: Contains only C and H, so it's a hydrocarbon. So, R 2.
S) C₂H₂: In C₂H₂, there are 2 carbon atoms and 2 hydrogen atoms. Moles of C = 2, Moles of H = 2. % C by mole = %. So, S 5.
Therefore, the correct match is P 3, Q 1, R 2, S 5.
Stream:JEESubject:ChemistryTopic:StoichiometrySubtopic:Calculation of Percentage yield
⏱ 2mℹ️ Source: QB
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