Hóa học

\(n_{Cu}=\dfrac{51,2}{64}=0,8\left(mol\right)\\ PTHH:2Cu+O_2\underrightarrow{t^o}2CuO\\ \left(mol\right)....0,8\rightarrow..0,4.....0,8\\ m_{CuO}=0,8.80=64\left(g\right)\)

a) PTHH :       \(S+O_2->SO_2\)

b) Ta có : \(n_S\) = \(\dfrac{m_S}{M_S}\) = 0.1 (mol)

Có :           \(n_S=n_{O_2}\)

           --> \(n_{O_2}\) = 0.1 (mol)

          => \(V_{O_2\left(đktc\right)}\) = \(n_{O_2}\) . 22.4 = 2.24 (L)

\(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\\ PTHH:S+O_2\underrightarrow{t^o}SO_2\\ \left(mol\right)..0,1\rightarrow0,1..0,1\\ V_{O_2}=0,1.22,4=2,24\left(l\right)\)

Đề có phải thiếu: "Còn lại là Cacbon"

a) \(PTHH:2SO_2+O_2\xrightarrow[V_2O_5]{450^oC}2SO_3\)

\(n_{SO_2}=\dfrac{32}{64}=0,5\left(mol\right)\\ n_{O_2}=\dfrac{10}{32}=0,3125\left(mol\right)\)

Lập tỉ lệ: \(\dfrac{n_{SO_2}}{2}< \dfrac{n_{O_2}}{1}\left(\dfrac{0,5}{2}< 0,3125\right)\)

=> SO2 hết O2 dư

Theo pt: \(n_{O_2\left(pư\right)}=\dfrac{n_{SO_2}.2}{3}=\dfrac{0,5.1}{2}=0,25\left(mol\right)\)

\(n_{O_2\left(dư\right)}=0,3125-0,25=0,0625\left(mol\right)\\ m_{O_2}=0,0625.32=2\left(g\right)\)

c) Theo pt, ta có:\(n_{SO_3}=n_{SO_2}=0,5\left(mol\right)\)

\(m_{SO_3}=0,5.80=40\left(g\right)\)

\(n_{CO_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

\(4X+O_2\underrightarrow{t^0}2X_2O\)

\(......0.1.....0.2\)

\(M_{X_2O}=\dfrac{18.8}{0.2}=94\left(\dfrac{g}{mol}\right)\)

\(\Leftrightarrow2X+16=94\)

\(\Leftrightarrow X=39\left(kali\right)\) 

Chúc bạn học tốt 

\(n_{Mg}=\dfrac{4.8}{24}=0.2\left(mol\right)\)

\(Mg+2H^+\rightarrow Mg^{2+}+H_2\)

\(0.2.......0.4....................0.2\)

\(V_{dd}=\dfrac{0.4}{1.5+0.5\cdot2}=0.16\left(l\right)\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(n_{Fe}=\dfrac{16.8}{56}=0.3\left(mol\right)\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0.1\left(mol\right)\)

\(2Fe+\dfrac{3}{2}O_2\underrightarrow{t^0}Fe_2O_3\)

\(0.2....0.15.........0.1\)

\(n_{Fe\left(pư\right)}=0.2\left(mol\right)< 0.3\Rightarrow Fedư\)

\(V_{O_2}=0.15\cdot22.4=3.36\left(l\right)\)

\(m_{Fe\left(dư\right)}=\left(0.3-0.2\right)\cdot56=5.6\left(g\right)\)

\(n_A = \dfrac{2,24}{22,4} = 0,1\ mol\\ M_A = \dfrac{3}{0,1} = 30(đvC)\\ \Rightarrow d_{A/O_2} = \dfrac{M_A}{M_{O_2}} = \dfrac{30}{32} = 0,9375 \)

nP = m/M = 12,4/31 = 0,4 (mol) 

Ta có PTHH:   4 P+  5 O2 --->  2 P2O5

Theo PT:        4    -   5           -   2                   (mol)

BC:                0.4  -    0.5       - 0.2                  (mol)

Suy ra:     V O2 = 22.4 x  n = 22.4 x 0.5 = 11.2 (l)

Suy ra:     m P2O5 = n x M = 0.2 x 142 = 28.4 (g)