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What volume of hydrogen will be liberated at NTP by the reaction of Zn on 50 ml of dilute H2SO4 of density 1.3/ml and having purity 40% ?
  • a)
    3.5 litre
  • b)
    5.94 litre
  • c)
    6.74 litre
  • d)
    none of these
Correct answer is option 'B'. Can you explain this answer?
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What volume of hydrogen will be liberated at NTP by the reaction of Zn...
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What volume of hydrogen will be liberated at NTP by the reaction of Zn...
To find the volume of hydrogen gas liberated, we need to follow the steps below:

Step 1: Calculate the number of moles of H2SO4
Given:
Volume of H2SO4 = 50 mL
Density of H2SO4 = 1.3 g/mL
Purity of H2SO4 = 40%

First, we need to calculate the mass of the H2SO4 using the density and volume:
Mass of H2SO4 = Volume of H2SO4 × Density of H2SO4
= 50 mL × 1.3 g/mL
= 65 g

Now, we can calculate the mass of pure H2SO4:
Mass of pure H2SO4 = Purity of H2SO4 × Mass of H2SO4
= 40% × 65 g
= 26 g

Next, let's calculate the number of moles of H2SO4 using its molar mass:
Molar mass of H2SO4 = 2(1 g/mol) + 32 g/mol + 4(16 g/mol)
= 98 g/mol

Number of moles of H2SO4 = Mass of H2SO4 / Molar mass of H2SO4
= 26 g / 98 g/mol
= 0.265 moles

Step 2: Calculate the number of moles of H2 gas produced
The balanced chemical equation for the reaction of Zn with H2SO4 is:
Zn + H2SO4 → ZnSO4 + H2

According to the equation, 1 mole of Zn reacts with 1 mole of H2SO4 to produce 1 mole of H2 gas. Therefore, the number of moles of H2 gas produced is also 0.265 moles.

Step 3: Convert moles of H2 gas to volume at NTP
At NTP (Normal Temperature and Pressure), 1 mole of any gas occupies a volume of 22.4 liters.

Volume of H2 gas at NTP = Number of moles of H2 gas × 22.4 liters/mole
= 0.265 moles × 22.4 liters/mole
= 5.94 liters

Therefore, the volume of hydrogen gas liberated at NTP by the reaction of Zn on 50 mL of dilute H2SO4 is 5.94 liters. Hence, the correct answer is option B.
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Read the passage given below and answer the following questions:Reductive alkylation is the term applied to the process of introducing alkyl groups into ammonia or a primary or secondary amine by means of an aldehyde or ketone in the presence of a reducing agent. The present discussion is limited to those reductive alkylations in which the reducing agent is hydrogen and a catalyst or "nascent" hydrogen, usually from a metalacid combination; most of these reductive alkylations have been carried out with hydrogen and a catalyst. The principal variation excluded is that in which the reducing agent is formic acid or one of its derivatives; this modification is known as the Leuckart reaction. The process of reductive alkylation of ammonia consists in the addition of ammonia to a carbonyl compound and reduction of the addition compound or its dehydration product. The reaction usually is carried out in ethanol solution when the reduction is to be effected catalytically:Since the primary amine is formed in the presence of the aldehyde it may react in the same way as ammonia, yielding an additional compound, a Schiff's base (RCH= NCH2R) and finally, a secondary amine. Similarly, the primary amine may react with the imine, forming an addition product which also is reduced to a secondary amine Finally, the secondary amine may react with either the aldehyde or the imine to give products which are reduced to tertiary amines.Similar reactions may occur when the carbonyl compound employed is a ketone.Q. The reaction of ammonia and its derivatives with aldehydes is called

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