Câu 1 : Gọi $n_C = a(mol) ; n_S = b(mol)$
Ta có : $12a + 32b = 5,6(1)$
$C + O_2 \xrightarrow{t^o} CO_2$
$S + O_2 \xrightarrow{t^o} SO_2$
$n_{O_2} = a + b = 0,3(2)$
Từ (1)(2) suy ra a = 0,2 ; b = 0,1
$m_C = 0,2.12 = 2,4(gam) ; m_S = 0,1.32 = 3,2(gam)$
Câu 2 :
$4R + 3O_2 \xrightarrow{t^o} 2R_2O_3$
Theo PTHH : $n_R = 2n_{R_2O_3} \Rightarrow \dfrac{2,16}{R} = 2.\dfrac{4,08}{2R + 16.3}$
$\Rightarrow R = 27$
Vậy R là nhôm, KHHH : Al
Câu 3 :
1 tấn= 1000 kg
$m_{Fe_3O_4} = 1000.90\% = 900(kg)$
$n_{Fe_3O_4} = \dfrac{900}{232}(kmol)$
$n_{Fe} = 3n_{Fe_3O_4} = \dfrac{900}{232}.3 = \dfrac{675}{58}(kmol)$
$m_{Fe} = \dfrac{675}{58}.56 = 651,72(kg)$
Câu 4 :
$n_{P_2O_5} = \dfrac{5,68}{142} = 0,04(mol) ; n_{H_2O} = \dfrac{2,7}{18} =0,15(mol)$
$P_2O_5 + 3H_2O \to 2H_3PO_4$
Ta thấy : $n_{P_2O_5} : 1 < n_{H_2O} : 3 $ nên $H_2O$ dư
$n_{H_3PO_4} = 2n_{P_2O_5} = 0,08(mol)$
$m_{H_3PO_4} = 0,08.98 = 7,84(gam)$
