Chemistry · Chemical equations and stoichiometry
Volume of in litres, liberated on heating 6.4 g of is:
Volume of \( N_{2}, \) in litres, liberated on heating 6.4 g of \( N H_{4} N O_{2} \) is:
- A. 44.8
- B. 22.4
- C. 11.2
- D. 2.24
Step-by-step solution
The balanced decomposition reaction is NH4NO2 → N2 + 2H2O. Molar mass of NH4NO2 = 64 g/mol, so 6.4 g corresponds to 0.1 mol. From stoichiometry, 0.1 mol NH4NO2 yields 0.1 mol N2. At STP, 1 mol of gas occupies 22.4 L, so volume of N2 = 0.1 × 22.4 = 2.24 L.
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