A certain barium halide exists as the hydrated salt BaX_2.2H_2 O, where X is the

Question

A certain barium halide exists as the hydrated salt BaX_2.2H_2 O, where X is the halogen. The barium content of the salt can be determined by gravimetric methods. A sample of the halide (0.2650 g) was dissolved in water (200 cm^3) and excess sulfuric acid added. The mixture was then heated and held at boiling for 45 minutes. The precipitate (barium sulfate) was filtered off, washed and dried. Mass of precipitate obtained = 0.2533 g. Determine the identity of X. A 0.4960 g sample of a CaCO_3 is dissolved in an acidic solution. The calcium is precipitated as CaC_2 O_4-H_2 O and the dry precipitate is found to weigh 0.6186 g. What is the percentage of CaO in the sample?

Explanation / Answer

Given :

BaX2.2H2O

X is halogen.

Sample of halide = 0.2650 g

Volume = 200 mL = 0.200 L

Mass of ppt = 0.2533 g

We know the ppt is of BaSO4

Lets find moles of BaSO4

= 0.2533 g / Molar mass of BaSO4

=0.2533 g / 233.43 g per mol

= 0.00108 mol BaSO4

Calculation of moles of halide

Moles X = Moles of BaSO4 x 2 mol halide (X) / 1 mol BaSO4

= 0.00108 mol BaSO4 * 2 mol halide (X) / 1 mol BaSO4

=0.00217 mol X

We know,

Number of moles of X = Mass in g / molar mass of X

Therefore Molar mass of X

= mass in g / moles of X

= 0.2650 g X / 0.00217 mol X

= 122.11 g/mol

This value is close to the mass of I

So the identity of the halide is Iodine

Q. 2 )

Given:

Mass of sample of CaCO3 = 0.4960 g

Formula for ppt : CaC2O4.H2O

Mass of ppt = 0.6186 g

Solution :

Percent of CaO = Mass of CaO / Total mass of sample ) x 100

Calculation of moles of CaC2O4.H2O

= 0.6186 g / 146.1123 g per mol

=0.004234 mol

We know moles ratio between CaO and CaC2H4.H2O is 1:1

So moles of CaO = 0.004234 mol

Mass of CaO = 0.004223 mol x molar mass

= 0.4223 mol x 56.0774 g per mol

= 0.237 g

Percent = (0.237 g/ 0.6186 g ) x 100

=38.4 %

Mass percent of CaO = 38.4 %

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