a) \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

b) \(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

c) \(M_A=1,172.29=34\left(g/mol\right)\)

\(n_A=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)

=> m_A = 1,5.34 = 51(g)

a)

$n_{O_2} = \dfrac{2,24}{22,4} = 0,1(mol)$
$m_{O_2} = 0,1.32 = 3,2(gam)$

b)
$n_{Cl_2} = \dfrac{7,1}{71} = 0,1(mol)$
$V_{Cl_2} = 0,1.22,4 = 2,24(lít)$

Câu 8:

\(d_{\dfrac{A}{KK}}>1\\ \Leftrightarrow M_A>M_{KK}\\ \Leftrightarrow M_A>29\\ Vậy:Chọn.A\)

(Vì 44>29>28>2)

\(Câu.7:C\\ Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\\ Câu.6:A\)

\(Câu.5:\\ n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{AlCl_3}=n_{Al}=0,3\left(mol\right)\\ m_{sp}=m_{AlCl_3}=0,3.133,5=40,05\left(g\right)\)

(Vì khí bay đi, thoát ra nên không tính trong KL Sp thu được)

a) \(n_{Al}=\dfrac{12,15}{27}=0,45\left(mol\right)\)

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

Xét tỉ lệ: \(\dfrac{0,45}{4}>\dfrac{0,3}{3}\)=> Al dư, O₂ hết

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

            0,4<--0,3-------->0,2

=> \(m_{Al\left(dư\right)}=\left(0,45-0,4\right).27=1,35\left(g\right)\)

b) \(m_{Al_2O_3}=0,2.102=20,4\left(g\right)\)

Câu 24: B

Câu 25: C

nO₂ = 8.96/22.4 =0.4 mol

nFe = 11.2 : 56 = 0.2 mol

PTHH: 3Fe + 2O_{2 ----->} Fe₂O₃

Theo PT: 3 - 2 -1 (mol)

BC:  0.2 - 0.4       (mol)

Ta có: \(\dfrac{3}{0.2}\)>\(\dfrac{2}{0.4}\)

Suy ra: số mol của Fe dư

\(n_{CH_4} = \dfrac{5,6}{22,4} = 0,25(mol)\\ CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ n_{H_2O} = n_{O_2} = 2n_{CH_4} = 0,5(mol)\\ \Rightarrow V_{O_2} = 0,5.22,4 = 11,2(lít) ; m_{H_2O} = 0,5.18 = 9(gam);\\ V_{kk} = 5V_{O_2} = 11,2.5 = 56(lít)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ m_{KMnO_4} = 0,5.2.158 = 158(gam)\\ 2KClO_3 \xrightarrow{t^o} 2KCl +3O_2\\ m_{KClO_3} = 0,5.\dfrac{2}{3}.122,5 = 40,83\ gam\)

\(n_{H_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\\ n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

LTL: \(\dfrac{1,5}{2}< 1,5\rightarrow O_2\) dư

Theo pt: \(\left\{{}\begin{matrix}n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}.1,5=0,75\left(mol\right)\\n_{H_2O}=n_{H_2}=1,5\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{O_2\left(dư\right)}=\left(1,5-0,75\right).32=24\left(g\right)\\V_{O_2}\left(1,5-0,75\right).22,4=16,8\left(l\right)\\m_{H_2O}=1,5.18=27\left(g\right)\end{matrix}\right.\)

\(n_{H_2}=n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(MOL\right)\) 
pthh: \(2H_2+O_2\underrightarrow{t^O}2H_2O\) 
LTL : \(\dfrac{1,5}{2}< \dfrac{1,5}{1}\) 
=> O2 dư , H2 hết 
theo pthh: nH2O = nH2 = 1,5 (mol) 
 => \(m_{H_2O}=1,5.18=27\left(g\right)\)