- 1(current)
Q. No.The mass of CO2 produced by heating 20 g of 20% pure limestone
as per the given below equation is:
\(\left[\mathrm{CaCO}_3 \stackrel{1200 \mathrm{~K}}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_2\right] \)
as per the given below equation is:
\(\left[\mathrm{CaCO}_3 \stackrel{1200 \mathrm{~K}}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_2\right] \)
| 1. | 1.32 g | 2. | 1.12 g |
| 3. | 1.76 g | 4. | 2.64 g |
Subtopic: Equation Based Problem |
69%
From NCERT
NEET - 2023
Please attempt this question first.
Hints
Please attempt this question first.
What mass of 95% pure CaCO3 will be required to neutralize 50 mL of 0.5 M HCl solution according to the following reaction?
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + 2H2O(l)
[Calculate upto the second place of decimal point]
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + 2H2O(l)
[Calculate upto the second place of decimal point]
| 1. | 9.50 g | 2. | 1.25 g |
| 3. | 1.32 g | 4. | 3.65 g |
Subtopic: Equation Based Problem |
51%
From NCERT
NEET - 2022
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NEET 2021 - Achiever Batch - Aryan Raj Singh
Hints
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NEET 2021 - Achiever Batch - Aryan Raj Singh
The number of moles of hydrogen molecules required to produce 20 moles of ammonia through Haber's process is -
| 1. | 40 mol | 2. | 10 mol |
| 3. | 20 mol | 4. | 30 mol |
Subtopic: Limiting Reagent | Equation Based Problem |
78%
From NCERT
NEET - 2019
To view explanation, please take trial in the course.
NEET 2021 - Achiever Batch - Aryan Raj Singh
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Links
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NEET 2021 - Achiever Batch - Aryan Raj Singh
- 1(current)
Q. No.
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