Bài 6:

$PTHH:2NaOH+H_2SO_4\to Na_2SO_4+2H_2O$

$n_{NaOH}=\dfrac{16}{40}=0,4(mol)$

Theo PT: $n_{H_2SO_4}=n_{Na_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,2(mol)$

$\Rightarrow m_{H_2SO_4}=0,2.98=19,6(g);m_{Na_2SO_4}=0,2.142=28,4(g)$

Bài 7:

$PTHH:CaCO_3\xrightarrow{t^o}CaO+CO_2\uparrow$

$n_{CaCO_3}=\dfrac{10}{100}=0,1(mol)$

Theo PT: $n_{CO_2}=n_{CaO}=n_{CaCO_3}=0,1(mol)$

$\Rightarrow m_{CaO}=0,1.56=5,6(g);V_{CO_2}=0,1.22,4=2,24(lít)$

Bài 8:

$PTHH:2Al(OH)_3+3H_2SO_4\to Al_2(SO_4)_3+6H_2O$

$n_{Al(OH)_3}=\dfrac{78}{78}=1(mol)$

Theo PT: $n_{H_2SO_4}=\dfrac{3}{2}n_{Al(OH)_3}=1,5(mol)$

$n_{Al_2(SO_4)_3}=\dfrac{1}{2}n_{Al(OH)_3}=0,5(mol)$

$\Rightarrow m_{H_2SO_4}=1,5.98=147(g)$

$m_{Al_2(SO_4)_3}=0,5.342=171(g)$

$\text{Đáp án+Giải thích các bước giải:}$

$\text{Bài 6:}$

$\text{PTHH: 2NaOH+H2SO4→Na2SO4+2H2O}$

$\text{Ta có: nNaOH=$\dfrac{16}{40}$= 0,4 mol}$

$\text{Theo PTHH ta có: nH2SO4=$\dfrac{1}{2}$×0,4= 0,2 mol}$

$\text{⇒mH2SO4=0,2×98= 19,6 g}$

$\text{Theo PTHH ta có: nNa2SO4=$\dfrac{1}{2}$×0,4= 0,2 mol}$

$\text{⇒mNa2SO4=0,2×142= 28,4 g}$

$\text{Bài 7:}$

$\text{PTHH: CaCO3→CaO+CO2}$

$\text{Ta có: nCaCO3=$\dfrac{10}{100}$= 0,1 mol}$

$\text{Theo PTHH ta có: nCO2=nCaCO3= 0,1 mol}$

$\text{⇒VCO2=0,1×22,4= 2,24 lít}$

$\text{Theo PTHH ta có: nCaO=nCaCO3= 0,1 mol}$

$\text{⇒mCaO=0,1×56= 5,6 g}$

$\text{Bài 8:}$

$\text{PTHH: 2Al(OH)3+3H2SO4→Al2(SO4)3+6H2O}$

$\text{Ta có: nAl(OH)3=$\dfrac{78}{78}$= 1 mol}$

$\text{Theo PTHH ta có: nH2SO4=$\dfrac{3}{2}$×1= 1,5 mol}$

$\text{⇒mH2SO4=1,5×98= 147 g}$

$\text{Theo PTHH ta có: nAl2(SO4)3=$\dfrac{1}{2}$×1= 0,5 mol}$

$\text{⇒mAl2(SO4)3=0,5×342= 171 g}$