step by step explanation please A solid sample of calcium carbonate is dissolved

Question

step by step explanation please

A solid sample of calcium carbonate is dissolved with hydrochloric acid (see unbalanced reaction below). The carbon dioxide produced from the reaction is all collected in a 2.50 L cylinder, at 25.0 degree C. If the reaction is performed with 23.07 g of CaCO_3 and 175 mL or 2.40 M HCI, what will be the pressure of the CO_2 in the cylinder? CaCO_3(s) + HCI(aq) rightarrow CaCI_2(aq) + H_2O(I) + CO_2(g) (a) 4.62 atm (b) 2.06 atm (c) 2.25 atm (d) 0.388 atm (e) 1.00 atm

Explanation / Answer

CaCO3 moles = mass / Molar mass of CaCO3

          = 23.07 g / 100 g/mol = 0.2307 mol

HCl moles = M x V ( inL) = 2.4 x 0.175 = 0.42 mol

as per reaction CaCO3 and HCl react in 1:2 ratio ( CaCO3 + 2HCl --> CaCl2 + 2H2O )

HCl moles needed for 0.23 CaCO3 moles = 2 x 0.23 0.46 but we have only 0.42 mol HCl

Hence HCl being relatively is low it is limiting reagent

CO2 moles produced = (1/2) HCl moles = 0.42/2 = 0.21

V = 2.5 L , T = 25C = 25+273 = 298 K

we use PV = nRT equation to find P

P x 2.5 L = 0.21 mol x 0.08206 liter atm/molK x 298 K

P = 2.05 atm

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