a, Chất rắn là MgO
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
0,8<-------------0,8<-------0,4
\(m=0,8.23+8=26,4\left(g\right)\)
\(b,m_{dd}=0,8.23+200-0,4.2=217,6\left(g\right)\\ \rightarrow C\%_{NaOH}=\dfrac{0,8.40}{217,6}.100\%=14,7\%\)
Tk:
a)2Al+ 6HCl→ 2AlCl3 +3H2↑
0,1________________0,15
Mg+ 2HCl→ MgCl2+ H2↑
0,2_______________0,2
MgO+ 2HCl→MgCl2+H2O
2Al+ 2NaOH+2H2O→ 2NaAlO2+ 3H2↑
0,1____________________________0,15
nHCl pư= 0,5.2.100110 =0,91 mol
nMgO=0,91−0,1.3−0,2.22= 0,105 mol
⇒ a= 0,1.27+0,2.24+ 0,105.40=11,7 g
b)
Dd B gồm:_______HCl dư ______AlCl3______MgCl2
_________________0,09________ 0,1_________0,305
NaOH+ HCl→ NaCl+ H2O
0,09 ___0,09
2NaOH+ MgCl2→ Mg(OH)2↓+ 2NaCl
0,61 ___0,305
3NaOH+ AlCl3→ 3NaCl+ Al(OH)3↓
0,3______ 0,1
VNaOH=0,09+0,61+0,32=0,5l
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
chất rắn thu đc là MgO (không pư)
m chất rắn là 8g => mMgO = 8(g)
\(pthh:Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
0,8 0,8 0,4
\(m_{Na}=0,8.23=18,4\left(g\right)\\ m_{hh}=18,4+8=26,4\left(g\right)\)
\(m_{\text{dd}}=18,6+200-\left(0,4.2\right)=217,8\left(g\right)\\ C\%_{NaOH}=\dfrac{0,8.40}{217,8}.100\%=14,692\%\)
