ib e :)

cô ơi em nghĩ cô nên sử dụng latex để các bạn dễ hiểu hơn ạ!

$a)PTHH:2KClO_3\xrightarrow{t^o}2KCl+3O_2$

$n_{O_2}=\dfrac{9}{32}=0,28125(mol)$

$\Rightarrow n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,1875(mol)$

$\Rightarrow m_{KClO_3}=0,1875.122,5=22,96875(g)$

$b)$ Theo PT: $n_{KCl}=n_{KClO_3}=0,1875(mol)$

$\Rightarrow m_{KCl}=0,1875.74,5=13,96875(g)$

$V_{hh}=22,4(0,25+0,1+0,45)=17,92(lít)$

Đặt CTHH của HC là $Cu_xS_yO_z$

$\%m_O=100-40-20=40\%$

$\Rightarrow x:y:z=\dfrac{40\%}{64}:\dfrac{20\%}{32}:\dfrac{40\%}{16}=1:1:4$

$\Rightarrow CTHH:CuSO_4$

$a)$ Tạo kết tủa trắng

$b)PTHH:Na_2SO_4+Ba(NO_3)_2\to 2NaNO_3+BaSO_4\downarrow$

$c)$ natri sunfat + bari nitrat ----> bari sunfat + natri nitrat

$4Na+O_2\xrightarrow{t^o}2Na_2O(1)$

$4Al+3O_2\xrightarrow{t^o}2Al_2O_3(2)$

$n_{Na}=\dfrac{4,6}{23}=0,2(mol);n_{O_2}=\dfrac{6,16}{22,4}=0,275(mol)$

$\Rightarrow n_{O_2(1)}=0,25.n_{Na}=0,05(mol)$

$\Rightarrow n_{O_2(2)}=0,275-0,05=0,225(mol)$

$\Rightarrow n_{Al_2O_3}=\dfrac{2}{3}n_{O_2(2)}=0,15(mol)$

$\Rightarrow m_{Al_2O_3}=0,15.102=15,3(g)$

$a)4P+5O_2\xrightarrow{t^o}2P_2O_5$

$n_P=\dfrac{4,65}{31}=0,15(mol)$

$\Rightarrow n_{O_2}=0,1875(mol)$

$\Rightarrow m_{O_2}=0,1875.32=6(g)$

$b)4Al+3O_2\xrightarrow{t^o}2Al_2O_3$

$n_{Al}=\dfrac{6,75}{27}=0,25(mol)$

$\Rightarrow n_{O_2}=0,1875(mol)$

$\Rightarrow m_{O_2}=0,1875.32=6(g)$

$c)2H_2+O_2\xrightarrow{t^o}2H_2O$

$n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)$

$\Rightarrow n_{O_2}=0,075(mol)$

$\Rightarrow m_{O_2}=0,075.32=2,4(g)$

$n_{H_2}=\dfrac{7,437}{24,79}=0,3(mol)$

$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2$

$\Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2}=0,2(mol)$

$\Rightarrow \%m_{Al}=\dfrac{0,2.27}{24,6}.100\%=21,95\%$

$\Rightarrow \%m_{Cu}=100-21,95=78,05\%$

$b)n_{AlCl_3}=n_{Al}=0,2(mol)$

$\Rightarrow m_{AlCl_3}=0,2.133,5=26,7(g)$

$c)n_{HCl}=3n_{Al}=0,6(mol)$

$\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,3}=2M$