Bảo toàn khối lượng : 

$m_{CO_2} = 12 - 7,6 = 4,4(gam)$
$n_{CaO} = n_{CaCO_3\ pư} = n_{CO_2} = \dfrac{4,4}{44} = 0,1(mol)$

$H = \dfrac{0,1.100}{12}.100\% = 83,33\%$

$\%m_{CaO} = \dfrac{0,1.56}{7,6}.100\% = 73,68\%$
$\%m_{CaCO_3} = 100\% -73,68\% = 26,32\%$

\(n_{CaCO_3}=\dfrac{12}{100}=0,12\left(mol\right)\\ PTHH:CaCO_3\underrightarrow{to}CaO+CO_2\\ x.........x........x\left(mol\right)\\ m_{rắn}=m_{CaCO_3\left(còn\right)}+m_{CaO}=\left(12-100x+56x\right)=7,6\\ \Leftrightarrow x=0,1\left(mol\right)\\ H=\dfrac{0,1}{0,12}.100\approx83,333\%\)

\(n_{CaCO_3}=\dfrac{400}{100}=4\left(mol\right)\\ CaCO_3\rightarrow\left(t^o\right)CaO+CO_2\uparrow\\ n_{CO_2\left(LT\right)}=n_{CaCO_3}=4\left(mol\right)\\ \Rightarrow n_{CO_2\left(TT\right)}=H.n_{CO_2\left(LT\right)}=75\%.4=3\left(mol\right)\\ \Rightarrow V_{CO_2\left(đktc\right)}=3.22,4=67,2\left(l\right)\)

CaCO3-> CaO +CO2

nCaCO3=200/100=2 (mol)

nCaO=nCaCO3=2 (mol)

khối lượng CaO sinh ra : mCaO = 2*56=112(g)

khối lượng CaCO3 dư : m CaCO3 dư= 134-112=22(g)

khối lương CaCO3 phản ứng : mCaCO3 pư= 200-22=178(g)

Hiệu suất H=178/200 *100%=89%

Bài 1 :

\(m_{giam}=m_{CO2}=400-312=88\left(g\right)\)

\(CaCO_3\underrightarrow{^{to}}CaO+CO_2\)

Theo lý thuyết ,

100g CaCO3 tạo 44g CO2

400g CaCO3 tạo 176g CO2

\(\Rightarrow H=\frac{88}{176}.100=50\%\)

Bài 2 :

a, \(n_{CaCO3}=2\left(mol\right)\)

\(H=75\%\Rightarrow\) Có 1,5mol CaCO3 phản ứng

\(CaCO_3\underrightarrow{^{to}}CaO+CO_2\)

1,5______1,5__________

A gồm 0,5mol CaCO3 ; 1,5mol CaO

\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

0,5______1_________0,5______0,5_________

\(CaO+2HCl\rightarrow CaCl_2+H_2O\)

1,5_____3_______1,5__________

\(\Rightarrow V=11,2\left(l\right)\)

b, \(n_{HCl}=4\left(mol\right)\)

\(\Rightarrow x=4.36,5:14,6\%=1000\left(g\right)\)

c,\(\Sigma n_{CaCl2}=2\left(mol\right)\Rightarrow m_{CaCl2}=222\left(g\right)\)

\(m_{dd_{spu}}=m_{CaCO3}+m_{CaO}+1000-m_{CO_2}\)

\(=1112\left(g\right)\)

\(\Rightarrow C\%_{CaCl2}=\frac{222.100}{1112}=19,96\%\)

mCaCO3 = 500*80%= 400 (g) 

nCaCO3 = 400/100 = 4 (mol) 

nCaCO3(pư) = 4*70%=2.8 (mol) 

CaCO3 -to-> CaO + CO2 

2.8..................2.8 

Chất rắn X : CaCO3 dư , CaO 

mX = ( 4 -2.8 ) *100 + 2.8*56 = 276.8 (g) 

%CaO = 2.8*56/276.8 * 100% = 56.64%

a)mCaCO3=500.80%=400(g) -> nCaCO3=400/100=4(mol)

PTHH: CaCO3 -to-> CaO + H2O

nCaO(LT)=nCaCO3=4(mol)

=> nCaO(TT)=4. 70%=2,8(mol)

=>mX=mCaO+ m(trơ)+ mCaCO3(chưa p.ứ)=2,8.56+100+ 1,2.100=376,8(g)

b) %mCaO= (156,8/376,8).100=41,614%

a)mCaCO3=500.80%=400(g) -> nCaCO3=400/100=4(mol)

PTHH: CaCO3 -to-> CaO + H2O

nCaO(LT)=nCaCO3=4(mol)

=> nCaO(TT)=4. 70%=2,8(mol)

=>mX=mCaO+ m(trơ)+ mCaCO3(chưa p.ứ)=2,8.56+100+ 1,2.100=376,8(g)

b) %mCaO= (156,8/376,8).100=41,614%

nCuO=0,02 mol

        CuO+H2=>H2O+Cu

Bđ0,02 mol

Pứ:x mol                      =>x mol

Dư:0,02-x mol

Cr sau pứ gồm Cu và CuO

=>64(0,02-x)+80x=1,344

=>16x=0,064=>x=0,004

H%=0,004/0,02.100%=20%

a) $n_{O_2} = 0,15(mol)$

\(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\)

0,3                         0,15           0,15       0,15                    (mol)

$H = \dfrac{0,15.158}{63,2}.100\% = 37,5\%$

b)

$m_B = 63,2 - 0,15.32 = 58,4(gam)$
$\%m_{K_2MnO_4} = \dfrac{0,15.197}{58,4}.100\% = 50,59\%$

$\%m_{MnO_2} = \dfrac{0,15.87}{58,4}.100\% = 22,35\%$
$\%m_{KMnO_4\ dư} = 100\% -50,59\% -22,35\% = 27,06\%$

\(Fe\left(NO_3\right)_2--->Fe_2O_3+NO_2+O_2\)

Bảo toàn nguyên tố Fe :

\(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(NO_3\right)_2}.90\%=\dfrac{1}{2}.\dfrac{23,4}{180}.90\%=0.065.90\%=0.0585\left(mol\right)\)

⇒ m_CR = 9.36(g)

Đặt \(n_{Fe}=x\left(mol\right)\)

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

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

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

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