Cây 23:
a) Gọi \(n_{Mg\left(NO_3\right)_2\left(pư\right)}=a\left(mol\right);n_{khí}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH: \(2Mg\left(NO_3\right)_2\xrightarrow[]{t^o}2MgO+4NO_2+O_2\)
a---------------->a---------->2a---->0,5a
=> 2a + 0,5a = 0,6 => a = 0,24 (mol)
Mà \(n_{Mg\left(NO_3\right)_2\left(bđ\right)}=\dfrac{37}{148}=0,25\left(mol\right)\)
=> \(H=\dfrac{0,24}{0,25}.100\%=96\%\)
b) Chất rắn X có: \(\left\{{}\begin{matrix}n_{Mg\left(NO_3\right)_2\left(dư\right)}=0,25-0,24=0,01\left(mol\right)\\n_{MgO}=0,24\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}m_{Mg\left(NO_3\right)_2}=0,24.40=9,6\left(g\right)\\m_{Mg\left(NO_3\right)_2\left(dư\right)}=0,01.148=1,48\left(g\right)\end{matrix}\right.\)
c) \(\overline{M}_Y=\dfrac{0,24.2.46+0,24.0,5.32}{0,6}=43,2\left(g/mol\right)\)
=> \(d_{Y/H_2}=\dfrac{43,2}{2}=21,6\)
d) \(4NO_2+O_2+2H_2O\rightarrow4HNO_3\)
Ban đầu: 0,48 0,12
Pư: 0,48 0,12
Sau pư: 0 0 0,48
=> \(C_{M\left(HNO_3\right)}=\dfrac{0,48}{2}=0,24M\)
