7,87 tấn quặng có \(m_{Al_2O_3}=7,87.60\%=4,722\left(tấn\right)\)

Ta có: Al₂O₃ ---> 2Al

          102(g)--->54(g)

       4,722(tấn)->2,5(tấn)

=> \(H=\dfrac{2}{2,5}.100\%=80\%\)

a) K₂O

PTK: 39.2 + 16 = 94 (đvC)

b) CaCl₂

PTK: 40 + 35,5.2 = 111 (đvC)

c) Al(NO₃)₃

PTK: 27 + (14 + 16.3).3 = 213 (đvC)

d) Ba(OH)₂

PTK: 137 + (16 + 1).2 = 171 (đvC)

e) Zn₃(PO₄)₂

PTK: 65.3 + (31 + 16.4).2 = 385 (đvC)

a) Gọi \(\left\{{}\begin{matrix}n_{NaNO_3}=a\left(mol\right)\\n_{Cu\left(NO_3\right)_2}=b\left(mol\right)\end{matrix}\right.\)

=> 85a + 188b = 4,43 (1)
PTHH: \(2NaNO_3\xrightarrow[]{t^o}2NaNO_2+O_2\)

              a-------------------------->0,5a

            \(2Cu\left(NO_3\right)_2\xrightarrow[]{t^o}2CuO+4NO_2+O_2\)

               b------------------------>2b------->0,5b

M_A = 19,5.2 = 39 (g/mol)
Áp dụng sơ đồ đường chéo, ta có:

\(\dfrac{n_{NO_2}}{n_{O_2}}=\dfrac{39-32}{46-39}=1\)

=> \(n_{NO_2}=n_{O_2}\)

=> 2b = 0,5a + 0,5b

=> 0,5a - 1,5b = 0 (2)

Từ (1), (2) => a = 0,03; b = 0,01

=> \(V_A=\left(0,5.0,03+0,01.2+0,01.0,5\right).22,4=0,896\left(l\right)\)

b) \(\left\{{}\begin{matrix}m_{NaNO_3}=0,03.85=2,55\left(g\right)\\m_{Cu\left(NO_3\right)_2}=0,01.188=1,88\left(g\right)\end{matrix}\right.\)

$\rm CaO + H_2O \rightarrow Ca(OH)_2$
$\rm P_2O_5 + 3H_2O \rightarrow 2H_3PO_4$
$\rm Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O$

PTHH: 2NaOH + H₂SO₄ ➝ Na₂SO₄ + 2H₂O
TPT:      2 mol       1 mol         1 mol
TĐB:    0,2 mol     0,1 mol      0,1 mol   

m_NaOH = \(\dfrac{m_{dd}}{100\%}.C\% = \dfrac{200}{100\%}.4\% = 8 (g)\)
n_NaOH = \(\dfrac{m}{M} = \dfrac{8}{40} = 0,2 (mol)\)
\(m_{H_2SO_4} = n.M = 0,1.98 = 9,8 (g)\)
\(m_{ddH_2SO_4} = \dfrac{m_{ct}.100\%}{C\%} = \dfrac{9,8.100\%}{9,8\%} = 100(g)\)
\(m_{ddNa_2SO_4} = m_{ddNaOH} + m_{ddH_2SO_4} = 200 + 100 = 300(g)\)
\(m_{Na_2SO_4} = n.M = 0,1.142 =14,2(g)\)
\(C\%_{Na_2SO_4} = \dfrac{m_{ct}}{m_{dd}}.100\% = \dfrac{14,2}{300}.100\% \) ≈ 4,7%