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a.\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{5,4}{27}=0,2mol\)
\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)
4 3 2 ( mol )
0,2 0,1
\(m_{Al_2O_3}=n_{Al_2O_3}.M_{Al_2O_3}=0,1.102=10,2g\)
b.\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2mol\)
\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{2,24}{22,4}=0,1mol\)
\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)
4 3 2
0,2 < 0,1 ( mol )
0,1 1/15
\(m_{Al_2O_3}=n.M=\dfrac{1}{15}.102=6,8g\)
a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH : 4Al + 3O2 -> 2Al2O3
0,2 0,1
\(m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)
b. \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH : 4Al + 3O2 -> 2Al2O3
0,1 \(\dfrac{0,2}{3}\)
Xét tỉ lệ : \(\dfrac{0,2}{4}>\dfrac{0,1}{3}\) => Al dư , O2 đủ
\(m_{Al_2O_3}=\dfrac{0,2}{3}.102=6,8\left(g\right)\)
2KMnO4-to>K2MnO4+MnO2+O2
0,3-----------------0,15-----0,15------0,15 mol
n KMnO4=\(\dfrac{47,4}{158}\)=0,3 mol
=>mcr=0,15.197.0,15.87=42,6g
=>VO2=0,15.22,4=3,36l
b) 4P+5O2-to>2P2O5
0,1--------------0,05
nP=\(\dfrac{3,1}{31}\)=0,1 mol
->O2 dư
=>m P2O5=0,05.142=7,1g
mKMnO4 = 47,4/158 = 0,3 (mol)
PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2
Mol: 0,3 ---> 0,15 ---> 0,15 ---> 0,15
m = 0,15 . 197 + 0,15 . 87 = 85,2 (g)
V = VO2 = 0,15 . 22,4 = 3,36 (l)
nP = 3,1/31 = 0,1 (mol)
PTHH: 4P + 5O2 -> (t°) 2P2O5
LTL: 0,1/4 < 0,15/5 => O2 dư
nP2O5 = 0,1/2 = 0,05 (mol)
mP2O5 = 0,05 . 142 = 7,1 (g)
Bài 2:
a) 2Mg + O2 --to--> 2MgO
b) \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PTHH: 2Mg + O2 --to--> 2MgO
_______0,2->0,1------>0,2
=> VO2 = \(\dfrac{0,1.0,082.\left(273+25\right)}{0,99}=2,468\left(l\right)\)
c) mMgO = 0,2.40 = 8(g)
Bài 3
a) Theo ĐLBTKL: mMg + mO2 = mMgO (1)
b) (1) => mMgO = 2,4 + 1,6 = 4(g)
c) \(nO_2=\dfrac{1,6}{32}=0,05\left(mol\right)\)
=> Số phân tử O2 = 0,05.6.1023 = 0,3.1023
\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2mol\)
\(PTHH:2H_2+O_2\underrightarrow{t^o}2H_2O\)
\(2\) \(:\) \(1\) \(:\) \(2\) \(\left(mol\right)\)
\(0,2\) \(:\) \(0,1\) \(:\) \(0,2\) \(\left(mol\right)\)
\(m_{H_2O}=n.M=0,2.\left(1.2+16\right)=3,6g\)
nH2 = 4,48 : 22,4 = 0,2 (mol)
nO2 = 11,2 : 22,4 = 0,5 (mol)
pthh : 2H2 + O2 -t--> 2H2O
LTL :
0,2/2 < 0,5/1
=> O2 du
theo pthh : nH2O = nH2 = 0,2 (mol)
=> mH2O = 0,2 . 18 = 8,6 (G)
nP = 3,1/31 = 0,1 (mol)
PTHH: 4P + 5O2 -t°-> 2P2O5
0,1---> 0,125--->0,05
VO2 = 0,125 . 22,4 = 2,8 (l)
mP2O5 = 0,05 . 142 = 7,1 (g)
\(n_P=\dfrac{3,1}{31}=0,1mol\)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
0,1 0,125 0,05
\(V_{O_2}=0,125\cdot22,4=2,8l\)
\(m_{P_2O_5}=0,05\cdot142=7,1g\)
Bài 1:
a, 2Mg + O2 \(\rightarrow\) 2MgO
b và c, Theo ĐLBTKL, ta có:
mMg + m\(O_2\) = mMgO
\(\Rightarrow m_{O_2}=8-4,8=3,2g\)
Bài 2:
a, Zn + 2HCl \(\rightarrow\) ZnCl2 + H2
b và c, Theo ĐLBTKL, ta có:
mZn + mHCl = m\(ZnCl_2\) + m\(H_2\)
\(\Rightarrow m_{HCl}=\left(27,2+0,4\right)-13=14,6g\)
a) PTHH: \(Zn+\dfrac{1}{2}O_2\xrightarrow[]{t^o}ZnO\)
b) Ta có: \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\) \(\Rightarrow n_{O_2}=0,15\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\)
c) PTHH: \(KClO_3\xrightarrow[MnO_2]{t^o}KCl+\dfrac{3}{2}O_2\uparrow\)
Theo PTHH: \(n_{KClO_3}=0,1\left(mol\right)\) \(\Rightarrow m_{KClO_3}=0,1\cdot122,5=12,25\left(g\right)\)
a. \(n_{Fe}=\dfrac{11.2}{56}=0,2\left(mol\right)\)
PTHH : 3Fe + 2O2 ---to---> Fe3O4
0,2 \(\dfrac{0.4}{3}\)
b. \(V_{O_2}=\dfrac{0.4}{3}.22,4=\dfrac{8.96}{3}\left(l\right)\)
c. PTHH : 2KClO3 -> 2KCl + 3O2
\(\dfrac{0.8}{3}\) \(\dfrac{0.4}{3}\)
\(m_{KClO_3}=\dfrac{0.8}{3}.122,5=\dfrac{98}{3}\left(g\right)\)
1. \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
\(2Mg+O_2\xrightarrow[]{t^\circ}2MgO\)
0,2 → 0,1
\(\Rightarrow V_{O_2}=0,1\cdot22,4=2,24\left(l\right)\)
2. \(n_{P_2O_5}=\dfrac{28,2}{142}=\dfrac{141}{710}\left(mol\right)\)
\(4P+5O_2\xrightarrow[]{t^\circ}2P_2O_5\)
\(\dfrac{141}{355}\) ← \(\dfrac{141}{710}\)
\(\Rightarrow m_P=\dfrac{141}{355}\cdot31\approx12,312\left(g\right)\)
1, \(2Mg+O_2--->2MgO\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
Theo pt \(n_{O_2}=\dfrac{n_{Mg}}{2}=\dfrac{0,2}{2}=0,1\left(mol\right)\)
\(V_{O_2}=n\times22,4=0,1\times22,4=2,24\left(lít\right)\)
2, \(4P+5O_2-->2P_2O_5\)
\(n_{P_2O_5}=\dfrac{28,2}{2\times31+5\times16}\approx0,2\left(mol\right)\)
Theo phương trình :
\(n_P=\dfrac{n_{P_2O_5}}{2}=\dfrac{0,2}{2}=0,1\left(mol\right)\)
\(m_P=0,1\times31=3,1\left(g\right)\)