Câu 8:

Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{0,2}{1}\), ta được O₂ dư.

Theo PT: \(\left\{{}\begin{matrix}n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{H_2}=0,05\left(mol\right)\\n_{H_2O}=n_{H_2}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{O_2\left(dư\right)}=0,15\left(mol\right)\)

\(\Rightarrow V_{O_2\left(dư\right)}=0,15.22,4=3,36\left(l\right)\)

\(m_{H_2O}=0,1.18=1,8\left(g\right)\)

Bạn tham khảo nhé!

Câu 9:

a, PT: \(2R+O_2\underrightarrow{t^o}2RO\)

Theo ĐLBT KL, có: m_R + m_O2 = m_RO

⇒ m_O2 = 4,8 (g)

\(\Rightarrow n_{O_2}=\dfrac{4,8}{32}=0,15\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,15.22,4=3,36\left(l\right)\)

b, Theo PT: \(n_R=2n_{O_2}=0,3\left(mol\right)\)

\(\Rightarrow M_R=\dfrac{19,2}{0,3}=64\left(g/mol\right)\)

Vậy: M là đồng (Cu).

Câu 10:

Ta có: m_BaCl2 = 200.15% = 30 (g)

a, m _dd  = 200 + 100 = 300 (g)

\(\Rightarrow C\%_{BaCl_2}=\dfrac{30}{300}.100\%=10\%\)

⇒ Nồng độ dung dịch giảm 5%

b, Ta có: \(C\%_{BaCl_2}=\dfrac{30}{150}.100\%=20\%\)

⇒ Nồng độ dung dịch tăng 5%.

Bạn tham khảo nhé!

a. \(n_{H_2}=\dfrac{1.12}{22,4}=0,05\left(mol\right)\)

\(n_{O_2}=\dfrac{2.24}{22,4}=0,1\left(mol\right)\)

PTHH : 2H₂ + O₂ ---t^o---> 2H₂O

            0,05   0,025           0,05

b. Ta thấy : 0,05 < 0,1 => H₂ đủ , O₂ dư

\(V_{O_2\left(dư\right)}=\left(0,1-0,025\right).22,4=1,68\left(l\right)\)

c. \(m_{H_2O}=0,05.18=0,9\left(g\right)\)

Ta có 

nH2 = 1,12 / 22,4 = 0,05 ( mol )

nO2 = 0,1 ( mol )

      2H2 + O2 ---to ---> 2H2O

có : 0,05    0,1

pư : 0,05   0,025          0,05

dư : 0        0,075

Xét tỉ lệ : 0,05 / 2 < 0,1 / 1 , ta được O2 dư

=> nO2 dư = 0,1 - 0,025 = 0,075 ( mol )

=> V = 0,075 . 22,4 = 1,68 ( l )

mH2O = 0,05 . 18 = 0,9 ( g )

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

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) 
 => \(m_{H_2O}=1,5.18=27\left(g\right)\)

\(n_{C_2H_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

\(n_{O_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)

      \(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{t^0}2CO_2+H_2O\)

\(Bđ:0.3.......0.5\)

\(Pư:0.2........0.5.........0.4.........0.2\)

\(Kt:0.1..........0..........0.4...........0.2\)

\(V_{CO_2}=0.4\cdot22.4=8.96\left(l\right)\)

\(V_{C_2H_2\left(dư\right)}=0.1\cdot22.4=2.24\left(l\right)\)

a, PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

Ta có: \(n_{H_2}=n_{O_2}=\dfrac{6,1975}{24,79}=0,25\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,25}{2}< \dfrac{0,25}{1}\), ta được O₂ dư.

Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{H_2}=0,125\left(mol\right)\)

\(\Rightarrow n_{O_2\left(dư\right)}=0,25-0,125=0,125\left(mol\right)\)

\(\Rightarrow V_{O_2\left(dư\right)}=0,125.24,79=3,09875\left(l\right)\)

b, Theo PT: \(n_{H_2O}=n_{H_2}=0,25\left(mol\right)\)

\(\Rightarrow m_{H_2O}=0,25.18=4,5\left(g\right)\)

c, PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{1}{6}\left(mol\right)\)

\(\Rightarrow m_{KClO_3}=\dfrac{1}{6}.122,5\approx20,42\left(g\right)\)

\(n_{O_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

\(n_{N_2O_5}=\dfrac{10.8}{108}=0.1\left(mol\right)\)

\(2N_2+5O_2\underrightarrow{t^0}2N_2O_5\)

\(.......0.25.....0.1\)

\(m_{O_2\left(dư\right)}=\left(0.3-0.25\right)\cdot32=1.6\left(g\right)\)

\(n_{Mg}=\dfrac{2.4}{24}=0.1\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(0.1....................................0.1\)

\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(2H_2+O_2\underrightarrow{^{t^0}}2H_2O\)

\(0.1.....0.05\)

\(m_{O_2\left(dư\right)}=\left(0.5-0.05\right)\cdot32=14.4\left(g\right)\)

\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{2,4}{22,4}\approx0,11\left(mol\right)\)

\(n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{1,6}{22,4}\approx0,07\)

\(2H_2+O_2\rightarrow2H_2O\)

2 mol-1mol---2 mol

Ta có: \(\dfrac{n_{H_2}}{2}=\dfrac{0,11}{2}\)

\(\dfrac{n_{O_2}}{1}=\dfrac{0,07}{1}\)

\(\Rightarrow\dfrac{n_{H_2}}{2}< \dfrac{n_{O_2}}{1}\)

Vậy \(O_2\) dư

Số mol O₂ dư:

\(n_{O_2}=\dfrac{0,07.1}{2}=0,035\left(mol\right)\)

Khối lượng O₂ dư

\(m_{O_2}=0,035.32=1,12\left(g\right)\)

Khối lượng nước thu được:

\(n_{H_2O}=\dfrac{0,07.2}{2}=0,07\left(mol\right)\)

\(\Rightarrow m_{H_2O}=n_{H_2O}.M_{H_2O}=0,07.18=1,26\left(g\right)\)

PT: \(2Mg+O_2\underrightarrow{t^o}2MgO\)

Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,25}{1}\), ta được O₂ dư.

Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{Mg}=0,1\left(mol\right)\)

\(\Rightarrow n_{O_2\left(dư\right)}=0,25-0,1=0,15\left(mol\right)\)

Bạn tham khảo nhé!

PTHH: \(2Mg+O_2\underrightarrow{t^o}2MgO\)

Ta có: \(\left\{{}\begin{matrix}n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,25}{1}\) \(\Rightarrow\) Magie p/ứ hết, Oxi còn dư

\(\Rightarrow n_{O_2\left(dư\right)}=0,25-0,1=0,15\left(mol\right)\)