Question

Question 3 5 pts What volume of carbon dioxide (in liters) at 1.10 atm and 37°C are produced when 33.7 g of limestone (CaCO3) undergoes completion? Balanced equation: CaCO315) + 2 HCl(aq) +CaCl2(aq) + CO2(g) + H20) 929L 3.37 L none of these options 7.79L

Answer 1

Answer :- Option (d) 7.79 L

solution :-

the reaction is given below

CaCO3(s) + 2 HCl(aq) +CaCl2(aq) + CO2(g) + H20)

No of moles = mass / molar mass

Molar mass of CaCO3 = 100.0869 g/mol

No of moles of CaCO3 = 33.7 g / 100.0869

No of moles of CaCO3 = 0.33671 moles of CaCO3

0.33671 moles of CaCO3 produces 0.33671 moles of CO2

since we know that

from ideal gas equation,

PV = nRT

so ,

V = nRT / P ................. eqn 1

it is given that

n= 0.33671 moles

P = 1.10 atm

T= 273+37 = 310 K

R = 0.082057 Latm / mol K

So from equation 1

V = 0.33671 *0.082057*310 /1.10

V = 7.79 L

So the correct answer is Option (d) 7.79 L

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