SATHEE: Some Basic Concepts Of Chemistry Question 5

Some Basic Concepts Of Chemistry Question 5

Question: The minimum quantity in gram of $H_{2} S$ needed to precipitate 63.5 g of $C u^{2^{+}}$ will be $(C u^{2^{+}} + H_{2} S \to C u_{2} S + H_{2})$ Black

Options:

A) 63.5 g

B) 31.75 g

C) 34 g

D) 2 g

Show Answer

Answer:

Correct Answer: C

Solution:

$C u^{2 +} + H_{2} S \to C u_{2} S + H_{2}$ 63.5g 1 mole of $C u^{2 +}$ requires 1 mole of $H_{2} S$ Or 63.5 g of $C u^{2 +}$ requires 34 g of $H_{2} S$ [Molar mass of $H_{2} S$ =2+32=34g] So 1 mole $H_{2} S$ is required, i.e, 34 g

Some Basic Concepts Of Chemistry Question 5

Question: The minimum quantity in gram of $H_{2} S$ needed to precipitate 63.5 g of $C u^{2^{+}}$ will be $(C u^{2^{+}} + H_{2} S \to C u_{2} S + H_{2})$ Black

Options:

Answer:

Solution:

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Some Basic Concepts Of Chemistry Question 5

Question: The minimum quantity in gram of H2S needed to precipitate 63.5 g of Cu2+ will be (Cu2++H2S→Cu2S+H2) Black

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: The minimum quantity in gram of $H_{2} S$ needed to precipitate 63.5 g of $C u^{2^{+}}$ will be $(C u^{2^{+}} + H_{2} S \to C u_{2} S + H_{2})$ Black