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\(n_{P_2O_5}=\dfrac{m}{M}=\dfrac{28,4}{31\cdot2+16\cdot5}=0,2\left(mol\right)\)
\(PTHH:4P+5O_2-^{t^o}>2P_2O_5\)
tỉ lệ 4 : 5 : 2
n(mol) 0,4<---0,5<-----0,2 (mol)
\(m_P=n\cdot M=0,4\cdot31=12,4\left(g\right)\)
Ta có: \(n_S=\dfrac{9,6}{32}=0,3\left(mol\right)\)
PT: \(S+O_2\underrightarrow{t^o}SO_2\)
Theo PT: \(n_{SO_2}=n_S=0,3\left(mol\right)\)
\(\Rightarrow V_{SO_2}=0,3.24,79=7,437\left(l\right)\)
PTHH: 3Fe + \(2O_2\) --->\(Fe_3O_4\)
theo pt: 3_____2_____________1
theo đề: x______y_____________0.01
nFe3O4 là: 0.01mol
\Rightarrow nO2= 0.01*2/1=0.02 mol
VO2= 0.02*22.4=0.448l
b, PTHH : 2KMnO4 ----> K2MnO4 + MnO2 + O2
theo pt: 2__________1________1______1
theo đề: x___________________________0.02
=> n KMnO4= 0.02*2/1= 0.04 mol
=>mKMnO4= 0.04*158=6.32g
a. số mol của Fe3O4 là :
2.32 : 232 =0.01 mol
theo tỉ lệ mol ta có số mol của Fe là:
0.01 * 3 = 0.03 mol
khối lượng sắt là: 0.03*56=1.68g
số mol oxi là: 0.01*2=0.02mol
thể tích oxi là: 0.02*22.4= 0.448g
b. 2KMnO_4 ---> K2MnO4 + MnO2 + O2
---> nKMnO_4 = 2nO2 = 0,04 mol ---> mKMnO_4=0.04*158=6.32g
a, \(n_P=\dfrac{24,8}{31}=0,8\left(mol\right)\)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
\(n_{O_2}=\dfrac{5}{4}n_P=1\left(mol\right)\) \(\Rightarrow V_{O_2}=1.22,4=2,24\left(l\right)\)
b, \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,4\left(mol\right)\Rightarrow m_{P_2O_5}=0,4.142=56,8\left(g\right)\)
c, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
\(n_{KMnO_4}=2n_{O_2}=2\left(mol\right)\Rightarrow m_{KMnO_4}=2.158=316\left(g\right)\)
\(n_P=\dfrac{m}{M}=\dfrac{24,8}{31}=0,8\left(mol\right)\)
\(PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\)
4 5 2
0,8 1 0,4
\(a.V_{O_2}=n.24,79=1.24,79=24,79\left(l\right)\\ b.m_{P_2O_5}=n.M=0,4.\left(31.2+16.5\right)=56,8\left(g\right)\)
\(c.PTHH:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2\downarrow+O_2\uparrow\)
2 1 1 1
0,8 0,4 0,4 0,4
\(m_{KMnO_4}=n.M=0,8.\left(39+55+16.4\right)=126,4\left(g\right).\)
\(n_{CuO}=0,2\left(mol\right)\)
\(CuO+H_2\xrightarrow[]{t^\circ}Cu+H_2O \)
0,2 → 0,2
\(\Rightarrow m_{Cu}=0,2\cdot64=12,8\left(g\right)\)
\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
0,1 0,1
\(n_{CuO}=\dfrac{16}{160}=0,1\left(mol\right)\)
\(m_{Cu}=0,1.64=6,4\left(g\right)\)
a) 4P + 5O2 --to--> 2P2O5
b) \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
PTHH: 4P + 5O2 --to--> 2P2O5
0,2-->0,25------->0,1
=> mP2O5 = 0,1.142 = 14,2(g)
c) VO2 = 0,25.22,4 = 5,6(l)
a) \(3Fe+2O_2-t^o->Fe_3O_4\)
b) \(n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)\)
Theo pthh : \(n_{Fe_3O_4}=\frac{1}{3}n_{Fe_3O_4}=\frac{0,1}{3}\left(mol\right)\)
=> \(m_{Fe_3O_4}=232\cdot\frac{0,1}{3}\approx7,73\left(g\right)\)
c) Theo pthh : \(n_{O2\left(pứ\right)}=\frac{2}{3}n_{Fe}=\frac{0,2}{3}\left(mol\right)\)
=> \(n_{O2\left(can.dung\right)}=\frac{0,2}{3}\div100\cdot120=0,08\left(mol\right)\)
=> \(V_{O2\left(can.dung\right)}=0,08\cdot22,4=1,792\left(l\right)\)
PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
a, Theo PT: \(n_{O_2}=\dfrac{5}{4}n_P=0,125\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,125.22,4=2,8\left(l\right)\)
b, Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,05\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=0,05.142=7,1\left(g\right)\)
Bạn tham khảo nhé!
\(PTPU:4P+5O_2\rightarrow2P_2O_5\)(đk:\(t^o\))
4 : 5 : 2 (tỉ lệ mol)
0,4 0,2 (mol)
\(n_{P_2O_5}=\dfrac{m}{M}=\dfrac{28,4}{\left(2.31+16.5\right)}=0,2\left(mol\right)\)
\(m_P=n.M=0,4.3112,4\left(g\right)\)
0,4.3112,4
=>??