\(n_{CaCO_3}=\dfrac{1}{100}=0,01kmol\\ CaCO_3\xrightarrow[]{t^0}CaO+CO_2\\ n_{CaO\left(lt\right)}=n_{CaCO_3}=0,01kmol\\ m_{CaO\left(lt\right)}=0,01.56=0,56tấn\\ m_{CaO\left(tt\right)}=0,56.90:100=0,504tấn\)

\(1.\\ a)V_{CH_4}=1,5.24,79=37,185l\\ 2.\\ m_{ddKCl}=20+60=80g\\ C_{\%KCl}=\dfrac{20}{80}\cdot100\%=25\%\)

a. Oxide tác dụng với H2SO4 loãng: K₂O, P₂O₅, ZnO, CuO, BaO, Na₂O, Al₂O₃, FeO.

PTHH:

\(K_2O+H_2SO_4\rightarrow K_2SO_4+H_2O\)

\(P_2O_5+3H_2SO_4\rightarrow3SO_3+2H_3PO_4\)

\(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)

\(BaO+H_2SO_4\rightarrow BaSO_4+H_2O\)

\(Na_2O+H_2SO_4\rightarrow Na_2SO_4+H_2O\)

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\).

\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)

b. Oxide tác dụng với KOH loãng: BaO, Na₂O, Al₂O₃, FeO.

PTHH:

\(CuO+2KOH\rightarrow Cu\left(OH\right)_2+K_2O\)

\(BaO+2KOH\rightarrow Ba\left(OH\right)_2+K_2O\)

\(Na_2O+2KOH\rightarrow2NaOH+K_2O\)

\(Al_2O_3+2KOH\rightarrow2KAlO_2+H_2O\)

\(FeO+2KOH\rightarrow Fe\left(OH\right)_2+K_2O\)

ok bạn nhó

a) B là \(Al_2\left(SO_4\right)_3\), C là \(Cu\)

\(b)n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,2        0,3                0,1                  0,3

\(m_{hh}=0,2.27+3,2=8,6g\\ \%m_{Cu}=\dfrac{3,2}{8,6}\cdot100=37,21\%\\ \%m_{Al}=100-37,21=62,79\%\\ c)C_{M_{H_2SO_4}}=\dfrac{0,3}{0,25}=1,2M\)

thank

\(n_{NaOH}=\dfrac{200.12\%}{40}=0,6\left(mol\right)\\ 3NaOH+FeCl_3\rightarrow Fe\left(OH\right)_3+3NaCl\\ 2Fe\left(OH\right)_3\rightarrow\left(t^o\right)Fe_2O_3+3H_2O\\ A:Fe\left(OH\right)_3;B:NaCl\\ a,n_{Fe\left(OH\right)_3}=\dfrac{0,6}{3}=0,2\left(mol\right);n_{NaCl}=n_{NaOH}=0,6\left(mol\right)\\ m_A=m_{Fe\left(OH\right)_3}=107.0,2=21,4\left(g\right)\\ m_{ddB}=200+200-21,4=378,6\left(g\right)\\ C\%_{ddNaCl}=\dfrac{0,6.58,5}{378,6}.100\approx9,271\%\\ b,Đặt:x=n_{Fe\left(OH\right)_3\left(p.ứ\right)}=x\left(mol\right)\\ \Rightarrow m_{rắn}=m_{Fe\left(OH\right)_3\left(chưa.phân.huỷ\right)}+m_{Fe_2O_3}=\left(21,4-107x\right)+80x=16,08\\ \Leftrightarrow21,4-27x=16,08\\ \Leftrightarrow27x=5,32\Leftrightarrow x=\dfrac{5,32}{27}\Rightarrow\%m_{A\left(nung\right)}=\dfrac{\dfrac{5,32}{27}}{0,2}.100\approx98,519\%\)

câu b em làm theo hướng m rắn đó bao gồm cả oxit lẫn hidroxit dư

\(A/n_{Na_2SO_4}=0,1.2=0,2mol\\ n_{BaCl_2}=0,2.3=0,6mol\\ Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl\\ \Rightarrow\dfrac{0,2}{1}< \dfrac{0,6}{1}\Rightarrow BaCl_2.dư\\ Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl\)

0,2mol          0,2mol        0,2mol          0,4mol

\(m_{rắn}=m_{BaSO_4}=0,2.233=46,6g\\ B/C_{M_{NaCl}}=\dfrac{0,4}{0,1+0,2}=\dfrac{4}{3}M\\ C_{M_{BaCl_2}}=\dfrac{0,6-0,2}{0,1+0,2}=\dfrac{4}{3}M\)

\(n_{Na_2SO_4}=0,1.2=0,2\left(mol\right);n_{BaCl_2}=0,2.3=0,6\left(mol\right)\\ PTHH:Na_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2NaCl\\ Vì:\dfrac{0,2}{1}< \dfrac{0,6}{1}\rightarrow BaCl_2dư\\ n_{BaSO_4}=n_{BaCl_2\left(p.ứ\right)}=n_{Na_2SO_4}=0,2\left(mol\right)\\ a,m_{rắn}=m_{BaSO_4}=233.0,2=46,6\left(g\right)\)

b, Dung dịch sau phản ứng có: NaCl và BaCl₂ dư

\(n_{BaCl_2\left(dư\right)}=0,6-0,2=0,4\left(mol\right)\\ n_{NaCl}=0,2.2=0,4\left(mol\right)\\V_{ddsau}=0,1+0,2=0,3\left(mol\right)\\ C_{MddBaCl_2}=\dfrac{0,4}{0,3}=\dfrac{4}{3}\left(M\right);C_{MddNaCl}=\dfrac{0,4}{0,3}=\dfrac{4}{3}\left(M\right)\)

\(m_{ddKNO_3}=40+3600=3640g\\ C_{\%KNO_3}=\dfrac{40}{3640}\cdot100\%=1,1\%\)

_ _ _ _ _ _ _

\(3600g\rightarrow360g\\ m_{ddKNO_3}=40+360=400g\\ C_{\%KNO_3}=\dfrac{40}{400}\cdot100\%=10\%\)