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a)\(2Mg + O_2 \xrightarrow{t^o} 2MgO\)
b)
\(n_{Mg} = \dfrac{2,4}{24} = 0,1(mol)\)
Theo PTHH :
\(n_{O_2} = \dfrac{1}{2}n_{Mg} = 0,05(mol)\\ \Rightarrow V_{O_2} = 0,05.22,4 = 1,12(lít)\)
c)
\(n_{MgO} = n_{Mg} = 0,1(mol)\\ \Rightarrow m_{MgO} = 0,1.40 = 4(gam)\)
d)
\(V_{không\ khí} = 5V_{O_2} = 1,12.5 = 5,6(lít)\)
nMg = 9,6/24 = 0,4 (mol)
2Mg + O2 ---to---> 2MgO
0,4____0,2_________0,4
VO2(đktc) = 0,2.22,4 = 4,48(l)
mMgO = 0,4.40 = 16(g)
a) \(n_{O_2}=\dfrac{11,2.20\%}{22,4}=0,1\left(mol\right)\)
PTHH: 2Mg + O2 --to--> 2MgO
0,2<--0,1--------->0,2
=> mMg = 0,2.24 = 4,8 (g)
b) nMgO = 0,2.40 = 8 (g)
\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{11,2}{22,4}=0,5mol\)
PTHH:
\(2Mg+O_2->2MgO\)
2 : 1 : 2 mol
1 : 0,5 : 1 mol
\(m_{Mg}=n.M=1.24=24g\)
\(m_{MgO}=n.M=1.\left(24+16\right)=40g\)
nO2 = 11,2/22,4 = 0,5 (mol)
PTHH: 2Mg + O2 -> (t°) 2MgO
Mol: 1 <--- 0,5 ---> 1
mMg = 1 . 24 = 24 (g)
mMgO = 1 . 40 = 40 (g)
\(1,PTHH:2Mg+O_2\xrightarrow{t^o}2MgO\\ 2,m_{Mg}+m_{O_2}=m_{MgO}\\ 3,m_{O_2}=15-9=6(g)\)
\(n_{Mg}=\dfrac{4.8}{24}=0.2\left(mol\right)\)
\(2Mg+O_2\underrightarrow{t^0}2MgO\)
\(0.2.......0.1........0.2\)
\(V_{O_2}=0.1\cdot22.4=2.24\left(l\right)\)
\(m_{MgO}=0.2\cdot40=8\left(g\right)\)
a)
2Mg + O2 --to--> 2MgO
0,2----->0,1------>0,2 (mol)
nMg = 4,48/24 = 0,2 (mol)
=> VO2 = 0,1.22,4 = 2,24 (lít)
b) mMgO = 0,2.(24 + 16) = 8 (g)
Câu 1:
a) \(PTHH:2Mg+O_2\underrightarrow{t^o}2MgO\)
b)
\(n_{Mg}=\frac{m_{Mg}}{M_{Mg}}=\frac{48}{24}=2\left(mol\right)\)
\(n_{O_2}=\frac{m_{O_2}}{M_{O_2}}=\frac{32}{32}=1\left(mol\right)\)
Lập tỉ lệ: \(\frac{2}{2}=\frac{1}{1}\)
=> PỨ hết
Theo ĐLBTKL, ta có:
\(m_{Mg}+m_{O_2}=m_{MgO}\)
\(48+32=m_{MgO}\)
\(m_{MgO}=80\left(g\right)\)
Bài 2:
a) \(PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\)
b)
\(n_P=\frac{m_P}{M_P}=\frac{6,2}{31}=0,2\left(mol\right)\)
Theo PTHH, ta có:
\(n_{O_2}=\frac{5}{4}n_P=\frac{5}{4}.0,2=0,25\left(mol\right)\)
\(V_{O_2}=n_{O_2}.22,4=0,25.22,4=5,6\left(l\right)\)
\(V_{kk}=V_{O_2}.5=5,6.5=28\left(l\right)\)
c)
Cách 1:
\(m_{O_2}=n_{O_2}.M_{O_2}=0,25.32=8\left(g\right)\)
Theo ĐLBTKL, ta có:
\(m_P+m_{O_2}=m_{P_2O_5}\)
\(6,2+8=m_{P_2O_5}\)
\(m_{P_2O_5}=14,2\left(g\right)\)
Cách 2:
Theo PTHH, ta có:
\(n_{P_2O_5}=\frac{2}{4}n_P=\frac{1}{2}n_P=\frac{1}{2}.0,2=0,1\left(mol\right)\)
\(m_{P_2O_5}=n_{P_2O_5}.M_{P_2O_5}=0,1.142=14,2\left(g\right)\)
nMg = 1.2/24 = 0.05 (mol)
2Mg + O2 -to-> 2MgO
0.05__0.025____0.05
mMgO = 0.05*40 = 2 (g)
VO2 = 0.025*22.4 = 0.56(l)
a, \(2Mg+O_2\underrightarrow{^{t^o}}2MgO\)
\(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)
\(n_{O_2}=\dfrac{1}{2}n_{MgO}=0,025\left(mol\right)\Rightarrow V_{O_2}=0,025.22,4=0,56\left(l\right)\)
b, Có lẽ đề cho oxi tác dụng với hidro chứ không phải oxit bạn nhỉ?
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PT: \(2H_2+O_2\underrightarrow{^{t^o}}2H_2O\)
Xét tỉ lệ: \(\dfrac{0,15}{2}>\dfrac{0,025}{1}\), ta được H2 dư.
THeo PT: \(n_{H_2O}=2n_{O_2}=0,05\left(mol\right)\Rightarrow m_{H_2O}=0,05.18=0,9\left(g\right)\)
Tham khảo ( tự lèm)
Câu 1:
a) PTHH:2Mg+O2to→2MgOPTHH:2Mg+O2to→2MgO
b)
nMg=mMgMMg=4824=2(mol)nMg=mMgMMg=4824=2(mol)
nO2=mO2MO2=3232=1(mol)nO2=mO2MO2=3232=1(mol)
Lập tỉ lệ: 22=1122=11
=> PỨ hết
Theo ĐLBTKL, ta có:
mMg+mO2=mMgOmMg+mO2=mMgO
48+32=mMgO48+32=mMgO
mMgO=80(g)mMgO=80(g)
Bài 2:
a) PTHH:4P+5O2to→2P2O5PTHH:4P+5O2to→2P2O5
b)
nP=mPMP=6,231=0,2(mol)nP=mPMP=6,231=0,2(mol)
Theo PTHH, ta có:
nO2=54nP=54.0,2=0,25(mol)nO2=54nP=54.0,2=0,25(mol)
VO2=nO2.22,4=0,25.22,4=5,6(l)VO2=nO2.22,4=0,25.22,4=5,6(l)
Vkk=VO2.5=5,6.5=28(l)Vkk=VO2.5=5,6.5=28(l)
c)
Cách 1:
mO2=nO2.MO2=0,25.32=8(g)mO2=nO2.MO2=0,25.32=8(g)
Theo ĐLBTKL, ta có:
mP+mO2=mP2O5mP+mO2=mP2O5
6,2+8=mP2O56,2+8=mP2O5
mP2O5=14,2(g)mP2O5=14,2(g)
Cách 2:
Theo PTHH, ta có:
nP2O5=24nP=12nP=12.0,2=0,1(mol)nP2O5=24nP=12nP=12.0,2=0,1(mol)
mP2O5=nP2O5.MP2O5=0,1.142=14,2(g)
Câu 1:
\(n_{Mg}=\dfrac{48}{24}=2\left(mol\right)\\ n_{O_2}=\dfrac{32}{32}=1\left(mol\right)\\ PTHH:2Mg+O_2\underrightarrow{t^o}2MgO\\ LTL:\dfrac{2}{2}=1\Rightarrow pư.đủ\\ Theo.pt:n_{MgO}=n_{Mg}=2\left(mol\right)\\ m_{MgO}=2.40=80\left(g\right)\)
Câu 2:
\(a,n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\\ b,Theo.pt:n_{O_2}=\dfrac{5}{4}n_{O_2}=\dfrac{5}{4}.0,2=0,25\left(mol\right)\\ V_{O_2}=0,25.22,4=5,6\left(l\right)\\ V_{kk}=5,6.5=28\left(l\right)\)
c, Cách 1:
mO2 = 0,25 . 32 = 8 (g)
Áp dụng ĐLBTKL, ta có:
mP + mO2 = mP2O5
=> mP2O5 = 6,2 + 8 = 14,2 (g)
Cách 2:
\(Theo.pt:n_{P_2O_5}=\dfrac{1}{2}n_P=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{P_2O_5}=0,1.142=14,2\left(g\right)\)