Question
Balance $\text{P}+\text{HNO}_3\xrightarrow{ \ \ \ \ \ \ \ \ }\text{H}_3\text{PO}_4+\text{NO}_2+\text{H}_2\text{O}$ by oxidation number method.
Multiply P by l, HNO3 by 5, we get$\text{P}+5\text{HNO}_3\xrightarrow{ \ \ \ \ \ \ \ \ }\text{H}_3\text{PO}_4+5\text{NO}_2+\text{H}_2\text{O}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| Elements | $\Delta\text{H}_{1}$ | $\Delta{\text{H}}_{2}$ | $\Delta_{\text{eg}}\text{H}$ |
| I | 520 | 7300 | -60 |
| II | 419 | 3051 | -48 |
| III | 1681 | 3374 | -328 |
| IV | 1008 | 1846 | -295 |
| V | 2372 | 5251 | +48 |
| VI | 738 | 1451 | -40 |
$\text{NO}^+,\text{N}_2,\text{SnCl}_2,\text{NO}^-_2$
$\text{CO}_2\text{CCl}_4,\text{O}_3,\text{NO}^-_2$