For the reaction NH4NO2 \(\to\) N2 + 2H2O, calculate the volume of nitrogen that would be produced at S. T. P from 3.2 g of the dioxonitrate(iii)salt. (Relative atomic masses: N = 14, O = 16, H = 1)
2.24 dm 3
2.24 cm3
1. 123
1. 12dm 3
4. 48 dm 3
Correct answer is D
\(NH_{4}NO_{2}\) \(\to\) \(N_{2}\) + \(2H_{2}O\)
\(64g\) \(\to\) \(22.4dm^{3}\)
\(3.2g\) will give \(\frac{22.4}{64}\) \(\times\) \(\frac{3.2}{64}\)
= \(1.12dm^{3}\)