1
\(n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\\
pthh:2Mg+O_2\underrightarrow{t^o}2MgO\\
\)
0,25 0,125
\(V_{O_2}=0,125.22,4=2,8\left(l\right)\\
V_{KK}=2,8.5=14\left(l\right)\)
2
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\
pthh:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
0,15 0,15 0,15
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\\
m_{FeCl_2}=127.0,15=19,05\left(g\right)\\
pthh:2H_2+O_2\underrightarrow{t^o}2H_2O\)
0,15 0,15
\(m_{H_2O}=0,15.18=2,7\left(g\right)\)
\(3\\
n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\\
n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\\
pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\\
LTL:\dfrac{0,4}{1}< \dfrac{0,5}{1}\)
=> H2SO4 dư
\(n_{H_2SO_4\left(p\text{ư}\right)}=n_{Zn}=0,4\left(mol\right)\\
m_{H_2SO_4\left(d\right)}=\left(0,5-0,4\right).98=9,8\left(g\right)\)
\(n_{ZnSO_4}=n_{Zn}=0,4\left(mol\right)\\
m_{ZnSO_{\text{ 4}}}=0,4.136=161=64,4\left(g\right)\)
dung dịch làm Qùy tím không chuyển màu vì ZnSO4 là muối
4
\(n_{Na}=\dfrac{13,8}{23}=0,6\left(mol\right)\\ pthh:Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\uparrow\)
0,6 0,6 0,3
\(m_{\text{dd}}=13,8+300-\left(0,3.2\right)=313,2\left(g\right)\\
C\%_{NaOH}=\dfrac{0,6.40}{313,2}.100\%=7,66\%\)
5
\(n_{CaO}=\dfrac{2,24}{56}=0,04\left(mol\right)\\
pthh:CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
0,04 0,04
\(m_{\text{dd}}=2,24+87,76=90\left(g\right)\\
C\%_{Ca\left(OH\right)_2}=\dfrac{0,04.74}{90}.100\%=3,289\%\)
Giup voii