Question

Determine the % yield if 8.3 L of CO2 at 28°C and 750 mm Hg are produced from the
decomposition of 42.5 g of iron(II) carbonate

Answer 1

balanced equation is

FeCO3 ------> FeO + CO2

from this it is clear that one mole of FeCO3 is giving on emole of  CO2

no of moles of FeCO3 = weight of FeCO3 / molar mass of FeCO3

= 42.5 g / 115.85 g/mol

= 0.367 moles

find the no of moles of CO2 using PV = nRT formula

P = 750 mmHg convert in to at m = 0.987 atm

V = 8.3 L, T = 273+ 28 = 301K, R = gas constant = 0.0821 L tam/mol-K

n = no of moles

put all these values in the above equation

0.987 x 8.3 = n x 0.0821 x 301

n = 0.3315

now

% of yield = (product moles / reactant moles ) x 100

= (0.3315 / 0.367 ) x 100

= 90.33 %