What Is the Mass Percentage of Each Component in a SrCO3 and BaCO3 Mixture?

[suspenc3]

A certain mixture contains only SrCO3 and BaCO3. When 1.60g of this mixture is treated with hydrochloric acid, 0.421g of carbon dioxide gas is liberated. Calculate the mass percentage of each component in the mixture. Reaction of the carbonate with the acid:

MCO3(s)+2H(aq)+--->M+2(aq) + H2O(l) +CO2(g).

Im completely lost on this problem, its kinda the same as the last I guess.

Any help will be appreciated.

 

[siddharth]

What are your thoughts/ideas on this question? You need to show some work before you get help.

 

[Blueshift5]

I can understand how this problem may seem daunting, but let me break it down for you. The given reaction is showing the decomposition of a carbonate (MCO3) when treated with hydrochloric acid (2HCl). In this case, the carbonate is either strontium carbonate (SrCO3) or barium carbonate (BaCO3).

The first step to solving this problem is to write out the balanced chemical equation for the reaction. From the given information, we know that 1.60g of the mixture contains both SrCO3 and BaCO3, and that 0.421g of carbon dioxide gas (CO2) is produced. The balanced equation would be:

SrCO3(s) + 2HCl(aq) → SrCl2(aq) + H2O(l) + CO2(g)

BaCO3(s) + 2HCl(aq) → BaCl2(aq) + H2O(l) + CO2(g)

Next, we need to calculate the number of moles of CO2 produced. We can do this by using the ideal gas law, PV = nRT. Since the reaction takes place at standard temperature and pressure (STP), we can use the values of 1 atm for pressure and 273 K for temperature. The volume of CO2 produced is not given, but we can assume it to be equal to the volume of the container, which is usually 22.4 L for STP. So, we have:

n = (PV)/(RT) = (1 atm)(22.4 L)/(0.08206 L atm/mol K)(273 K) = 0.991 mol CO2

Now, we need to determine the number of moles of the mixture that was used. Since we have 1.60g of the mixture, we can assume that it contains 1.60g of each component (SrCO3 and BaCO3). The molar mass of SrCO3 is 147.63 g/mol and the molar mass of BaCO3 is 197.34 g/mol. So, we have:

n = (1.60 g)/(147.63 g/mol) = 0.01084 mol SrCO3

n = (1.60 g)/(197.34 g/mol) = 0.00811 mol BaCO3

Now, we can use the number