Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
b, Ta có: \(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{2}n_{O_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)
c, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Có: O2 hao hụt 40% → H% = 100 - 40 = 60%
Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,4\left(mol\right)\)
\(\Rightarrow n_{KMnO_4\left(TT\right)}=\dfrac{0,4}{60\%}=\dfrac{2}{3}\left(mol\right)\)
\(\Rightarrow m_{KMnO_4}=\dfrac{2}{3}.158\approx105,3\left(g\right)\)
a, \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
b, \(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)
Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,2\left(mol\right)\Rightarrow m_{P_2O_5}=0,2.142=28,4\left(g\right)\)
c, \(n_{O_2}=\dfrac{5}{4}n_P=0,5\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,5.22,4=11,2\left(l\right)\)
d, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{1}{3}\left(mol\right)\Rightarrow m_{KClO_3}=\dfrac{1}{3}.122,5=\dfrac{245}{6}\left(g\right)\)
a) Ta có PT:
4P + 5O2 ---> 2P2O5
b) n\(O_2\)= \(\frac{44,8}{22,4}\)= 2(mol)
Theo PT ta có:
n\(P_2O_5\)= \(\frac{2}{5}\)n\(O_2\)=\(\frac{2}{5}\).2=0,8(mol)
=> m\(P_2O_5\)=142. 0,8=113,6(g)
c) Ta có PT:
2KMnO4 ---> K2MnO4 + MnO2 + O2
Theo PT ta có:
n\(KMnO_4\)=2n\(O_2\)=4(mol)
=> m\(KMnO_4\)=158.4=632(g)
nO2 = \(\dfrac{1,12}{22,4}=0,05\) mol
Pt: 4P + 5O2 --to--> 2P2O5
.............0,05 mol--> 0,02 mol
mP2O5 thu được = 0,02 . 142 = 2,84 (g)
Pt: 2KMnO4 --to--> K2MnO4 + MnO2 + O2
....0,1 mol<-----------------------------------0,05 mol
mKMnO4 cần dùng = 0,1 . 158 = 15,8 (g)
nO2 = 0,05 mol
4P + 5O2 → 2P2O5
⇒ nP2O5 = 0,025 mol
⇒ mP2O5 = 0,025.142 = 3,55 (g)
2KMnO4 → K2MnO4 + MnO2 + O2
⇒ nKMnO4 = 0,1 mol
⇒ mKMnO4 = 0,1.158 = 15,8(g)
PTHH: \(4P+5O_2\rightarrow2P_2O_5\)
\(n_{O_2}=\dfrac{22.4}{22.4}=1\)
Cứ 5 mol \(O_2\) PƯ thu được 2 mol \(P_2O_5\)
__ 1__________________\(\dfrac{2}{5}\)_________
\(m_{P_2O_5}=142.\dfrac{2}{5}=56.8\left(g\right)\)
\(2KMnO_4\rightarrow K_2Mn_4+MnO_2+O_2\)
Cứ 2 Mol \(KMnO_4\) PƯ sinh ra 1 mol \(O_2\)
\(m_{KMnO_4}=158\times2=316\left(g\right)\)
(?)
Số mol oxi đã dùng ở dktc là: \(n_{O_2}=\dfrac{22.4}{22.4}=1\left(mol\right)\)
PTHH: \(5O_2+4P-^{\text{dđiều kiện nhiệt độ}}->2P_2O_5\)
Từ PTHH => \(n_{P_2O_5}=\dfrac{2}{5}n_{O_2}=0.4\left(mol\right)\)
Khối lượng P2O5 thu được sau phản ứng là:
\(m_{p_2O_5}=0,4.142=56,8\left(g\right)\)
PTHH:
\(2KMnO_4-^{\text{dđiều kiện nhiệt độ}}->K_2MnO_4+MnO_2+O_2\)
Từ PTHH => \(n_{KMnO_4}=2n_{O_2}=2\left(mol\right)\)
Khối lượng KMnO4 cần dùng là:
\(m_{KMnO_4}=2.158=316\left(g\right)\)
a.b.\(n_{Mg}=\dfrac{m}{M}=\dfrac{6,4}{24}=\dfrac{4}{15}mol\)
\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\)
4/15 2/15 ( mol )
\(V_{O_2}=n.22,4=\dfrac{2}{15}.22,4=\dfrac{224}{75}l\)
c.\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)
4/15 2/15 ( mol )
\(m_{KMnO_4}=n.M=\dfrac{4}{15}.158=\dfrac{632}{15}g\)
nMg = 6,4 : 24= 0,26(mol)
pthh : 2Mg+O2 -t--> 2MgO
0,26 --> 0,13 (mol )
=> VO2(đktc) = 0,13.22,4=2,912(l)
pthh : 2KMnO4-t--> K2MnO4 + MnO2+ O2
0,26<------------------------------0,13(mol)
=> mKMnO4 = 0,26.158= 41,08(g)
\(a,PTHH:4P+5O_2\xrightarrow{t^o}2P_2O_5\\ b,n_P=\dfrac{6,2}{31}=0,2(mol)\\ \Rightarrow n_{O_2}=\dfrac{5}{4}n_P=0,25(mol)\\ \Rightarrow V_{O_2(đktc)}=0,25.22,4=5,6(l)\\ c,n_{P_2O_5}=\dfrac{1}{2}n_P=0,1(mol)\\ \Rightarrow m_{P_2O_5}=0,1.142=14,2(g)\)
\(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\\ a,PTHH:4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ b,n_{O_2}=\dfrac{5}{4}.0,4=0,5\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,5.22,4=11,2\left(l\right)\\ c,n_{P_2O_5}=\dfrac{2}{4}.0,4=0,2\left(mol\right)\\ m_{P_2O_5}=142.0,2=28,4\left(g\right)\)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
\(nO_2=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(\Rightarrow nP_2O_5=\dfrac{2}{5}nO_2=\dfrac{2}{5}.0,1=0,04\left(mol\right)\)
\(\Rightarrow mP_2O_5=0,04.\left(31.2+16.5\right)=5,68\left(g\right)\)
c/
pthh: \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\uparrow\)
\(nO_2=0,1\left(mol\right)\)
\(\Rightarrow nKMnO_4=\dfrac{2}{1}.nO_2=\dfrac{2}{1}.0,1=0,2\left(mol\right)\)
\(\Rightarrow mKMnO_4=0,2.\left(39+55+16.4\right)=31,6\left(g\right)\)