. A mixture of Na2O and BaO weighing 6.0 g is treated with dil. H2SO4 and 6.0 g of
BaSO4 is obtained. What percentage of BaO is present in the sample ? (Ba = 137)
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Answers
Answer:
65.8 %
Explanation:
This Is a neutralisation reaction because an acid is reacting with a base this will form salt and water
Since Barium sulphate is formed this means that Barium oxide is the one which reacts we write a balanced chemocal equation of the reaction
H2SO4 + BaO –> BaSO4 + H2O
Since we have the mass of barium sulphate we can find its moles
Moles = mass/ molar mass
S =32
O = 16
Therefore the molar mass is
137 + 32 + 16× 4 = 233
6.0 ÷ 233 = 0.0258
The mole ratio of BaO to BaSO4 is 1:1 therefore they have the same moles
We can now find the mass of BaO on the mixture because we have the moles and molar mass
The molar mass is 137 + 16 = 153
Mass = moles × Molar mass
153 × 0.0258 = 3.95 g
We can now find the percentage of BaO in the sample
3.95 ÷ 6.0 × 100% = 65.8 %
Explanation:
here's your answer
65.8%
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