Question

Acid precipitation dripping on limestone produces carbon dioxide by the following reaction: 

15.5 mL of CO2 was produced at 25.0°C and 790.0 mm Hg. 

Part 1 How many moles of CO2 were produced? 

 Part 2 (1 point) How many milligrams of CaCO3, were consumed?  

Answer 1

Part 1

Answer

0.0006583mol

Explanation

Ideal gas equation is

PV = nRT

P = Pressure ,790mmHg = 1.039atm

V = Volume , 15.5ml = 0.0155L

n = number of moles, ?

R = gas constant , 0.082057(L atm/mol K)

T = Temperature , 25℃ = 298.15K

n = PV/RT

= 1.039atm × 0.0155L/ (0.082057(L atm /mol K) × 298.15K)

= 0.0006583mol

Part 2

Answer

65.89mg

Explanation

CaCO3(s) + 2H⁺(aq) -------> Ca²⁺(aq) + CO2(g) + H2O(l)

stoichiometrically, 1mole of CaCO3 gives 1mole of CO2

number of moles of CO2 produced = 0.0006583mol

Therefore

number of moles of CaCO3 consumed = 0.0006583mol

mass of CaCO3 consumed = 0.0006583mol × 100.09g/mol = 0.06589g = 65.89mg