Chemistry MCQ Practice

Zinc nitrate crystals decompose on heating according to the following equation:

 

 2Zn(NO₃)₂.6H₂O (s) → 2ZnO (s) + 4NO₂ (g) + O₂ (g) + 12H₂O (g)

 

If 14.85 g of zinc nitrate crystals are heated strongly and then allowed to cool, what is the total volume of gas obtained at room temperature and pressure?

 

A)  2.4 dm³

 

B)  3.0 dm³

 

C)  5.1 dm³

 

D)  10.2 dm³

 

 

Answer: B

 

- At room temp and pressure, water is no longer a gas
→ 2 moles of Zn(NO₃)₂.6H₂O will produce only 5 moles of gases.

 

M_r of Zn(NO₃)₂.6H₂O = 297
No. of moles of Zn(NO₃)₂.6H₂O = 14.85/297 = 0.05 mol
No. of moles of gases = 5/2 x 0.05 = 0.125 mol
Volume of gases = 0.125 x 24 = 3.0 dm³