$6,0\; g$ of a compound $ A $ of formula $ C_3H_6O_6$ are treated with an excess of sodium. We collect NTP $ 1.3 \; L $ of hydrogen at $ 25^oC $ and $ 0.9 \; atm $. Prove it can not be ethylmethylether ! First look for the number of moles of this compound and of hydrogen !
For answers, use (possibly multiple times) the arrows ↑ and ↓ at the top! Complete please this question before moving on to the next!
$n_A=\frac{m_A}{M_A}=\frac{6,0}{60}=0,1\; mol$ $n_{H_2}$ $=$ $\frac{P\cdot V}{R\cdot T}$ $=$ $\frac{0,95\cdot 1,3}{0,083\cdot 298}$ $=$ $0,05\; mol$
Reason !
Then one mole of the compound would release half a mole of hydrogen: It is necessary that the compound has a single unique $ H $ atom (half of a $ H_2 $ molecule !) releasable by sodium. This can only be the case for propan-1-ol or propan-2-ol and not for ether which has groups of 3, 3 or 2 equivalent $ H $ atoms!
