PTHH: CaCO₃ \(\underrightarrow{to}\) CaO + CO₂

Theo PT: 100g → 56g

Theo bài: 500g → x (g)

\(m_{CaO}lt=\dfrac{500\times56}{100}=280\left(g\right)\)

\(\Rightarrow H=\dfrac{m_{CaO}tt}{m_{CaO}lt}\times100\%=\dfrac{224}{280}\times100\%=80\%\)

Đặt \(n_{Fe}=x\left(mol\right)\)

Rắn gồm \(\left\{{}\begin{matrix}Fe\\Fe_2O_3\left(dư\right)\end{matrix}\right.\)

\(PTHH:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\\ \left(mol\right)....0,5x.........\leftarrow x\)

\(m_{Fe}+m_{Fe_2O_3\left(dư\right)}=m_{rắn}\\ \Leftrightarrow56x+\left(24-0,5x.160\right)=19,2\\ \Leftrightarrow56x+24-80x=19,2\\ \Leftrightarrow24x=4,8\\ \Leftrightarrow x=0,2\)

\(H=\dfrac{m_{Fe_2O_3\left(pư\right)}}{m_{Fe_2O_3}}.100\%=\dfrac{m_{Fe_2O_3}-m_{Fe_2O_3\left(dư\right)}}{m_{Fe_2O_3}}.100\%=\left(1-\dfrac{m_{Fe_2O_3\left(dư\right)}}{m_{Fe_2O_3}}\right).100\%=\left(1-\dfrac{24-0,5.0,2.160}{24}\right).100\%=\dfrac{200}{3}\approx66,67\%\)

Đáp án : B

n_HCHO = ¼ n_Ag = 0,03 mol

,n_{CH3OH bđ} = 0,0375 mol

=> H% = 80%

\(a,n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ PTHH:2K+2H_2O\rightarrow2KOH+H_2\uparrow\\ Theo.pt:n_K=2n_{H_2}=2.0,1=0,2\left(mol\right)\\ m_K=0,2.39=7,8\left(g\right)\\ m_{K_2O}=17,2-7,8=9,4\left(g\right)\\ b,n_{CuO\left(bđ\right)}=\dfrac{12}{80}=0,15\left(mol\right)\\ PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ LTL:0,15>0,1\Rightarrow Cu.dư\)

Gọi n_{CuO (pư)} = a (mol)

=> n_Cu = a (mol)

m_{chất rắn sau pư} = 80(0,15 - a) + 64a = 10,8

=> a = 0,075 (mol)

=> n_{H2 (pư)} = 0,075 (mol)

\(H=\dfrac{0,075}{0,1}=75\%\)

\(n_{CaCO_3}=\dfrac{20}{100}=0,2\left(mol\right)\)

PT: \(CaCO_3\underrightarrow{t^o}CaO+H_2O\)

\(n_{CaO\left(LT\right)}=n_{CaCO_3}=0,2\left(mol\right)\)

\(\Rightarrow m_{CaO\left(LT\right)}=0,2.56=11,2\left(g\right)\)

\(\Rightarrow H=\dfrac{8}{11,2}.100\%\approx71,43\%\)

H₂+CuO-to>Cu+H₂O

 0,22---------------0,22

n _H2=\(\dfrac{4,928}{22,4}\)=0,22 mol

n _CuO=\(\dfrac{20}{80}\)=0,25 mol

=>H2 hết , CuO dư

=>m _Cu =0,22.64=14,08g

=>H=\(\dfrac{12}{14,08}.100\)=85,23%

nH2 = 0,22 (mol)

nCuO = 20/80 = 0,25 (mol)

nCu (TT) = 12/64 = 0,1875 (mol)

PTHH: CuO + H2 -> (t°) Cu + H2O

LTL: 0,25 > 0,22 => CuO dư

nCu (LT) = nH2 = 0,22 (mol)

H = 0,1875/0,22 = 85,22%

\(n_{Al\left(OH\right)_3}=\dfrac{19.5}{78}=0.25\left(mol\right)\)

\(n_{Al\left(OH\right)_3\left(pư\right)}=a\left(mol\right)\)

\(2Al\left(OH\right)_3\underrightarrow{^{^{t^0}}}Al_2O_3+3H_2O\)

\(a...............0.5a\)

\(m_{Cr}=m_{Al_2O_3}+m_{Al\left(OH\right)_3\left(dư\right)}=102\cdot0.5a+19.5-78a=15.45\left(g\right)\)

\(\Leftrightarrow a=0.15\)

\(H\%=\dfrac{0.15}{0.25}\cdot100\%=60\%\)

Gọi $n_{CaCO_3} = a(mol)$

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

Theo PTHH : 

$n_{CO_2} = n_{CaCO_3\ pư} = 0,75a(mol)$
Bảo toàn khối lượng : 

$100a = 80 + 0,75a.44 \Rightarrow a = 1,194(mol)$
$m = 1,194.100 = 119,4(gam)$