Answer:28.605
Explanation:First, the molar mass of of SrCO3, BaCO3 and CO2 has to be calculated, (using the molar mass of each element Sr = 87.62, Ba = 137.327, C=12.011, O= 16.00)
The molar masses are;
SrCO3 = 87.62 + 12.011 + (3*16) = 147.631g/mol
BaCO3 = 79.904 + 12.011 + (3*16) = 197.34 g/mol
CO2 = 12.011 + (2*16) = 44.011 g/mol
To obtain one of the equations to solve the problem;
The sample is made of SrCO3 and BaCO3 and has a mass of 0.846 g. Representing the mass of SrCO3 as ma and that of BaCO3 as mb. The first equation can be written as:
ma + mb = 0.846g (1)
To obtain another equation in order to be able to determine the different percentages of the compounds (SrCO3 and BaCO3) that make of the sample, a relationship can be obtained by determining the relationship between the number of moles of CO2 formed as the mass of the SrCO3 and BaCO3;
The number of moles of CO2 formed = (mass of CO2)/(molar mass) =0.234/44.011 =0.00532moles
CO2 contains 1 mole of carbon (C) so therefore 0.00532 moles of CO2 contains 0.00532 moles of C
The sample produced 0.00532 moles of CO2, therefore the number of moles SrCO3 and BaCO3 that produced this amount can be calculated using the formula;
= (mass )/(molar mass)
No of moles of SrCO3 and BaCO3 will be ma/147.631 and mb/197.34 moles respectively
The total amount of C molecules produced by SrCO3 and BaCO3 will be 0.00532 moles of C
The second equation can be written as
ma/147.631 + mb/197.34= 0.00532 (2)
Solving Equation (1) and (2) simultaneously;
ma = 0.604g; mb = 0.242g
Therefore the percentage of BaCO3 = (mass of BaCO3 )/(mass of sample )*100
= 0.242/(0.846 )*100
= 28.605%
