1 mol [55 + 2 × 16 = 87 g] MnO2 reacts completely with 4 mol [4 × 36.5 = 146 g] of HCl.
∴ 5.0 g of MnO2 will react with = \(\frac {146\ g }{ 87 \ g} × 5.0\ g\) of HCl
= 8.4 g of HCl
Hence, 8.4 g of HCl will react completely with 5.0 g of manganese dioxide.
If 0.01 mol of $\mathrm{P_4O_{10}}$ is removed from 0.1 mol, then the remaining molecules of $\mathrm{P_4O_{10}}$ will be: