9. (10 pts) A mixture of MnSo, and MnSO4H,O has a mass of 2,005 g. After heating to drive off all the water the ma...

Question

Question

science chemistry

Answer 1

Answer #1

Mass of water loss = ( 2.005 - 1.780 ) g = 0.225 g.

Moles of water present = (mass of water/molar mass) = (0.225/18)

= 0.0125

Now,

Moles of MnSO₄ present in MnSO₄, 4 H₂O

= ( moles of water/4)

= (0.0125/4)

= 0.003125

Now, mass of MnSO₄ = 0.003125 × molar mass of MnSO₄ = 0.003125 (g) × 151 (g/mol)

= 0.4718 g.

Total mass of MnSO₄, 4 H₂O present in the mixture

= 0.4718 + 0.225

= 0.6968 g.

Therefore mass percent of MnSO₄, 4H₂O in the mixture

= mass of MnSO4,4H20 x 100 mass of micture

= 0.6968 x 100 2.005

= 34.75 %.

Answer 2

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