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How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3and NaHCO3 containing equimolar amounts of both?
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How many mL of 0.1 M HCl are required to react completely with 1 g mix...
Calculation of Required Volume of HCl
1. Molar Mass Calculation:
- Molar mass of Na2CO3 = 105.99 g/mol
- Molar mass of NaHCO3 = 84.01 g/mol
2. Equimolar Mixture Calculation:
- 1 g mixture contains equimolar amounts of Na2CO3 and NaHCO3
- Therefore, each component weighs 0.5 g
3. Moles of Na2CO3 and NaHCO3:
- Moles = mass / molar mass
- Moles of Na2CO3 = 0.5 g / 105.99 g/mol
- Moles of NaHCO3 = 0.5 g / 84.01 g/mol
4. Common factor for reaction with HCl:
- Na2CO3 + 2HCl → 2NaCl + H2O + CO2
- NaHCO3 + HCl → NaCl + H2O + CO2
5. Calculating Total Moles of HCl needed:
- Total moles of HCl required = Moles of Na2CO3 + Moles of NaHCO3
- Total moles of HCl required = (0.5 / 105.99) + (0.5 / 84.01)
6. Volume Calculation:
- Molarity (M) = moles / volume (L)
- Volume (L) = moles / Molarity
- Volume of 0.1 M HCl = Total moles of HCl required / 0.1
7. Converting Volume to mL:
- 1 L = 1000 mL
- Therefore, the calculated volume should be converted to mL
By following the above steps, you can determine the number of mL of 0.1 M HCl required to completely react with the 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both.
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How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3and NaHCO3 containing equimolar amounts of both?
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How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3and NaHCO3 containing equimolar amounts of both? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3and NaHCO3 containing equimolar amounts of both? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3and NaHCO3 containing equimolar amounts of both?.
How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3and NaHCO3 containing equimolar amounts of both? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3and NaHCO3 containing equimolar amounts of both? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3and NaHCO3 containing equimolar amounts of both?.
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