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Barium may be analyzed for by precipitating it as BaSO4 and weighing the precipitate formed. When a 0.2680 g sample of a barium compound was treated with excess H2SO4, 0.1060 g of BaSO4 was formed.

What is the percentage of barium in the compound? [S= 32.0 g; O = 16.0 g; Ba = 137.33 g]

Sagot :

Answer:

Explanation:

To find the percentage of barium in the compound, we can follow these steps:

   Calculate moles of BaSO4 formed:

   From the given data, 0.1060 g of BaSO4 was formed.

   Molar mass of BaSO4=137.33 g/mol (Ba)+32.06 g/mol (S)+4×16.00 g/mol (O)=233.39 g/mol

   Molar mass of BaSO4=137.33g/mol (Ba)+32.06g/mol (S)+4×16.00g/mol (O)=233.39g/mol

   Number of moles of BaSO4 formed:

   moles of BaSO4=0.1060 g233.39 g/mol=0.000454 mol

   moles of BaSO4=233.39g/mol0.1060g​=0.000454mol

   Determine moles of Ba in BaSO4:

   According to the balanced chemical equation for the reaction:

   Ba2++SO42−→BaSO4

   Ba2++SO42−​→BaSO4​

   1 mole of BaSO4 contains 1 mole of Ba.

   moles of Ba=0.000454 mol

   moles of Ba=0.000454mol

   Calculate the mass of Ba in the sample:

   Now, we find the mass of Ba in the sample that produced BaSO4:

   mass of Ba=moles of Ba×molar mass of Ba=0.000454 mol×137.33 g/mol=0.0623 g

   mass of Ba=moles of Ba×molar mass of Ba=0.000454mol×137.33g/mol=0.0623g

   Calculate the percentage of Ba in the compound:

   Finally, the percentage of Ba in the compound is:

   Percentage of Ba=(mass of Bamass of sample)×100%mass of sample=0.2680 g(given)Percentage of Ba=(0.0623 g0.2680 g)×100%=23.25%

   Percentage of Ba=(mass of samplemass of Ba​)×100%mass of sample=0.2680g(given)Percentage of Ba=(0.2680g0.0623g​)×100%=23.25%

Therefore, the percentage of barium in the compound is ( \boxed{

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