a) \(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)

PTHH: \(4P+5O_2\xrightarrow[]{t^o}2P_2O_5\)

           0,4-->0,5----->0,2

b) \(V_{O_2}=0,5.24,79=12,395\left(l\right)\)

c) \(m_{P_2O_5}=0,2.142=28,4\left(g\right)\)

d) PTHH: \(2Cu+O_2\xrightarrow[]{t^o}2CuO\)

                  1<----0,5

`=> m_{Cu} = 1.64 = 64 (g)`

\(m_2=\dfrac{15}{30\%}=50\left(g\right)\\ m_1=50-15=35\left(g\right)\)

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

\(PTHH:Fe_2O_3+3H_2\xrightarrow[]{}2Fe+3H_2O\)

tỉ lệ          :1              3          2         3

số mol     :0,1           0,3       0,2      0,3

\(m_{Fe_2O_3}=0,1.160=16\left(g\right)\)

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

\(Fe_2O_3+3H_2\xrightarrow[]{t^\circ}2Fe+3H_2O\uparrow\)

  0,1          ←       0,2

\(\Rightarrow m_{Fe_2O_3}=0,1\cdot160=16\left(g\right)\)

a, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

____0,1_____0,2______0,1_____0,1 (mol)

\(V_{H_2}=0,1.24,79=2,479\left(l\right)\)

\(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)

b, \(C_{M_{HCl}}=\dfrac{0,2}{0,1}=1\left(M\right)\)

c, \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)

PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Xét tỉ lệ: \(\dfrac{0,15}{1}>\dfrac{0,1}{1}\), ta được CuO dư.

Theo PT: \(n_{CuO\left(pư\right)}=n_{Cu}=n_{H_2}=0,1\left(mol\right)\Rightarrow n_{CuO\left(dư\right)}=0,05\left(mol\right)\)

⇒ m _{chất rắn} = m_{CuO (dư)} + m_Cu = 0,05.80 + 0,1.64 = 10,4 (g)

a)

Làm giấy quỳ tím hóa xanh là dung dịch Bazo

\(CaO+H_2O->Ca\left(OH\right)_2\)

b)

Làm giấy quỳ tím hóa đỏ là dung dịch axit

\(SO_3+H_2O->H_2SO_4\)

c)

\(Zn+2HCl->ZnCl_2+H_2\)

d)

\(2KClO_3->2KCl+3O_2\)

\(n_{Fe_2o_3}=\dfrac{36}{160}=0,2\left(m\right)\)

\(PTHH:Fe_2O_3+3H_2\xrightarrow[]{}2Fe+3H_2O\)

tỉ lệ          :1             3           2         3

số mol     :0,2          0,6        0,4       0,6

\(m_{Fe}=0,4.56=22,4\left(g\right)\)

b)\(V_{H_2}=0,6.22,4=13,44\left(l\right)\)

Câu 9:

Ta có: \(m_{BaCl_2}=200.26\%=52\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{52}{208}=0,25\left(mol\right)\)

\(n_{H_2SO_4}=0,3.1=0,3\left(mol\right)\)

PT: \(BaCl_2+H_2SO_4\rightarrow2HCl+BaSO_{4\downarrow}\)

Xét tỉ lệ: \(\dfrac{0,25}{1}< \dfrac{0,3}{1}\), ta được H₂SO₄ dư.

Theo PT: \(\left\{{}\begin{matrix}n_{HCl}=2n_{BaCl_2}=0,5\left(mol\right)\\n_{BaSO_4}=n_{H_2SO_4\left(pư\right)}=n_{BaCl_2}=0,25\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,05\left(mol\right)\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05.98=4,9\left(g\right)\)

\(m_{BaSO_4}=0,25.233=58,25\left(g\right)\)

\(m_{HCl}=0,5.36,5=18,25\left(g\right)\)

Câu 10:

\(n_{NaOH}=\dfrac{40}{40}=1\left(mol\right)\)

\(\Rightarrow C_{M_{NaOH}}=\dfrac{1}{0,234}\approx4,2735\left(M\right)\)

Câu 11:

\(n_{HCl\left(2M\right)}=0,7.2=1,4\left(mol\right)\)

\(n_{HCl\left(3M\right)}=1,1.3=3,3\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{1,4+3,3}{0,7+1,1}\approx2,611\)