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student determines the manganese(II) content of a solution by first precipitating it as manganese(II) hydroxide, and then decomposing the hydroxide to manganese(II) oxide by heating. How many grams of manganese(II) oxide should the student obtain if his solution contains 54.0 mL of 0.491 M manganese(II) nitrate?

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Answer:

1.88 g of manganese(II) oxide

Explanation:

We could exprese the reactions that take place as:

Mn²⁺ (aq) + 2OH⁻ (aq) → Mn(OH)₂ (s)

Mn(OH)₂ → MnO + H₂O

The important aspect is that the ratio of Mn²⁺ to Mn(OH)₂ and then to MnO remains as 1.

  • Now we can calculate the starting moles of Mn²⁺ with the concentration (0.491 M) and the volume (54.0 mL = 0.054 L)

0.491 M * 0.054 L =0.026514 mol Mn²⁺

Then we calculate the moles of MnO and finally its mass, using its molecular weight:

0.026514 mol Mn²⁺ * [tex]\frac{1molMn(OH)_{2}}{1molMn^{2+}}* \frac{1molMnO}{1molMn(OH)_{2}} *\frac{70.94g}{1molMnO} =[/tex] 1.88 g MnO

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