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One natural source of atmospheric CO2 is precipitation reaction such as the precipitation of silicates in the oceans (up to the closest integer):
Mg2+ (aq) + SiO2 (dispersed) + 2HCO3(aq) → MgSiO3(s) + 2CO2(g) + H2O(I)
How many grams of magnesium silicate would be precipitated during the formation of 100 L of CO2 at 30°C and 775 torr?
(Mg = 24, Si = 28, C = 12, O = 16)
Correct answer is '205'. Can you explain this answer?
Verified Answer
One natural source of atmospheric CO2 is precipitation reaction such a...
Moles of CO2 formed = n = PV/RT
PV = 775/760 atm x 100L
RT- 0.0821 L atm mol-1 K-1 x 303 K
n = 4.0992
If 2 moles CO2 are formed then MgSiO3 precipitated is 1 mol and when 4.0992 moles CO2 and formed MgSiO3 would be
= 4.0992/2 = 2.0496 mol (molar mass of MgSiO= 100 g mol-1
= 2.0496 x 100 g = 205g
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One natural source of atmospheric CO2 is precipitation reaction such a...

Given data:
- Volume of CO2 produced = 100 L
- Temperature (T) = 30°C = 303 K
- Pressure (P) = 775 torr
- Molar masses: Mg = 24 g/mol, Si = 28 g/mol, C = 12 g/mol, O = 16 g/mol

Calculations:
1. Convert temperature to Kelvin: T = 30°C + 273 = 303 K
2. Convert pressure to atm: P = 775 torr / 760 = 1.018 atm
3. Use ideal gas law to find moles of CO2 produced:
PV = nRT
n = PV / RT = 1.018 atm * 100 L / 0.0821 L.atm/mol.K * 303 K = 5 mol
4. From the balanced equation, 1 mol of MgSiO3 produces 2 mol of CO2:
Moles of MgSiO3 = 5 mol / 2 = 2.5 mol
5. Calculate the molar mass of MgSiO3:
Mg: 1 * 24 g/mol = 24 g/mol
Si: 1 * 28 g/mol = 28 g/mol
O: 3 * 16 g/mol = 48 g/mol
Total molar mass = 24 + 28 + 48 = 100 g/mol
6. Calculate the mass of MgSiO3 precipitated:
Mass = 2.5 mol * 100 g/mol = 250 g

Therefore, approximately 250 grams of magnesium silicate would be precipitated during the formation of 100 L of CO2 at 30°C and 775 torr.
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One natural source of atmospheric CO2 is precipitation reaction such as the precipitation of silicates in the oceans (up to the closest integer):Mg2+ (aq) + SiO2 (dispersed) + 2HCO3(aq) → MgSiO3(s) + 2CO2(g) + H2O(I)How many grams of magnesium silicate would be precipitated during the formation of 100 L of CO2 at 30°C and 775 torr?(Mg = 24, Si = 28, C = 12, O = 16)Correct answer is '205'. Can you explain this answer?
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One natural source of atmospheric CO2 is precipitation reaction such as the precipitation of silicates in the oceans (up to the closest integer):Mg2+ (aq) + SiO2 (dispersed) + 2HCO3(aq) → MgSiO3(s) + 2CO2(g) + H2O(I)How many grams of magnesium silicate would be precipitated during the formation of 100 L of CO2 at 30°C and 775 torr?(Mg = 24, Si = 28, C = 12, O = 16)Correct answer is '205'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about One natural source of atmospheric CO2 is precipitation reaction such as the precipitation of silicates in the oceans (up to the closest integer):Mg2+ (aq) + SiO2 (dispersed) + 2HCO3(aq) → MgSiO3(s) + 2CO2(g) + H2O(I)How many grams of magnesium silicate would be precipitated during the formation of 100 L of CO2 at 30°C and 775 torr?(Mg = 24, Si = 28, C = 12, O = 16)Correct answer is '205'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One natural source of atmospheric CO2 is precipitation reaction such as the precipitation of silicates in the oceans (up to the closest integer):Mg2+ (aq) + SiO2 (dispersed) + 2HCO3(aq) → MgSiO3(s) + 2CO2(g) + H2O(I)How many grams of magnesium silicate would be precipitated during the formation of 100 L of CO2 at 30°C and 775 torr?(Mg = 24, Si = 28, C = 12, O = 16)Correct answer is '205'. Can you explain this answer?.
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