calculate the mass % caco3 in the 103 g mixture containing caco3 and na2co3
Answers
Answer:(2)48.5%
Explanation:If we add actual mass of both compounds tnen we got 206g but we have 103 g
Then it means we have 0.5 mole of each.
In 0.5 mole,mass of caco3= 50g
Mass% of caco3= Mass of caco3/ total mass ×100% =50/103×100%
By solving we get
48.5% ( approx)
Answer:
The answer is B) 48.5%. This is calculated by finding the molar mass of CaCO₃ (100.08 g/mol) and Na₂CO₃ (106.00 g/mol).
Given:
The mass of the mixture of calcium carbonate and sodium carbonate is 103g.
To find:
The mass percentage of CaCO₃ in a 103g mixture of Na₂CO₃.
Solution:
The total molar mass of the mixture is 206.08 g/mol. The mass of CaCO₃ in the mixture is then calculated by multiplying the molar mass of CaCO₃ by the moles of CaCO₃ produced from the reaction, divided by the total molar mass of the mixture. The moles of CaCO₃ produced from the reaction can be calculated using the ideal gas law and the data are given.
The moles of CaCO₃ produced from the reaction are calculated as follows:
n = PV/RT
n = (24 L * 1atm)/(0.0821 L*atm/K*mol * 300K) = 0.087 mol
The mass of CaCO₃ in the mixture is then calculated by multiplying the molar mass of CaCO₃ by the moles of CaCO₃ produced from the reaction, divided by the total molar mass of the mixture:
Mass of CaCO₃ in the mixture = (100.08 g/mol * 0.087 mol)/206.08 g/mol = 0.049 g
The mass percentage of CaCO₃ in the mixture is then calculated by dividing the mass of CaCO₃ in the mixture by the total mass of the mixture, multiplied by 100:
Mass percentage of CaCO₃ in the mixture = (0.049 g/103 g) * 100 = 48.5%
Therefore, the percentage mass of CaCO₃ is 48.5%.
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