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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:
\(Theo.pt:n_{P_2O_5}=\dfrac{1}{2}n_P=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{P_2O_5}=0,1.142=14,2\left(g\right)\)
a, - Phần 1: \(2Na+2H_2O\rightarrow2NaOH+H_2\)
- Phần 2: \(4Na+O_2\underrightarrow{t^o}2Na_2O\)
\(2Mg+O_2\underrightarrow{t^o}2MgO\)
b, Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{Na}=2n_{H_2}=0,2\left(mol\right)\)
\(\left\{{}\begin{matrix}n_{Na_2O}=\dfrac{1}{2}n_{Na}=0,1\left(mol\right)\\n_{MgO}=n_{Mg}\end{matrix}\right.\)
\(\Rightarrow0,1.62+40n_{Mg}=12,2\Rightarrow n_{Mg}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Na}=\dfrac{0,2.23}{0,2.23+0,15.24}.100\%\approx56,1\%\\\%m_{Mg}\approx43,9\%\end{matrix}\right.\)
\(1,PTHH:2Mg+O_2\xrightarrow{t^o}2MgO\\ 2,m_{Mg}+m_{O_2}=m_{MgO}\\ 3,m_{O_2}=15-9=6(g)\)
\(m_{Mg}+m_{O_2}\rightarrow m_{MgO}\Leftrightarrow4,8g+m_{O_2}\rightarrow8\Leftrightarrow m_{O_2}=3,2g\)
1. a. \(PTHH:2Mg+O_2\overset{t^o}{--->}2MgO\left(1\right)\)
b. Ta có: \(n_{Mg}=\dfrac{24}{24}=1\left(mol\right)\)
Theo PT(1): \(n_{O_2}=\dfrac{1}{2}.n_{Mg}=\dfrac{1}{2}.0,1=0,5\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,5.22,4=11,2\left(lít\right)\)
c. \(PTHH:2KClO_3\xrightarrow[t^o]{MnO_2}2KCl+3O_2\left(2\right)\)
Theo PT(2): \(n_{KClO_3}=\dfrac{2}{3}.n_{O_2}=\dfrac{2}{3}.0,5=\dfrac{1}{3}\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=\dfrac{1}{3}.122,5=40,83\left(g\right)\)
2. \(PTHH:3Fe+2O_2\overset{t^o}{--->}Fe_3O_4\)
Ta có: \(n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01\left(mol\right)\)
a. Theo PT: \(n_{Fe}=3.n_{Fe_3O_4}=0,01.3=0,03\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,03.56=1,68\left(g\right)\)
b. Theo PT: \(n_{O_2}=2.n_{Fe_3O_4}=2.0,01=0,02\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,02.22,4=0,448\left(lít\right)\)
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)
Áp dụng định luật BTKL:
\(a,m_{Mg}+m_{O_2}=m_{MgO}\\ b,m_{O_2}=m_{MgO}-m_{Mg}=50-42=8\left(g\right)\)
nMg = 5,76/24 = 0,24 (mol)
PTHH: 2Mg + O2 -> (t°) 2MgO
nMgO = 0,24 (mol)
mMgO = 0,24 . 40 = 9,6 (g)
nMg = 5,76 : 24 = 0,24 ( mol )
pthh : 2Mg+ O2 -t--> 2MgO
0,24->0,12-->0,24 (mol)
=> m = mMgO = 0,24 . 40 = 9,6 (g)
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: \(n_{H_2O}=2n_{O_2}=0,05\left(mol\right)\Rightarrow m_{H_2O}=0,05.18=0,9\left(g\right)\)
\(a,PTHH:2Mg+O_2\underrightarrow{t^o}2MgO\\ b,n_{MgO}=\dfrac{m}{M}=\dfrac{8}{40}=0,2\left(mol\right)\\ Theo.PTHH:n_{Mg}=n_{MgO}=0,2\left(mol\right)\\ m_{Mg}=n.M=0,2.24=4,8\left(g\right)\)
\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)
a) PTHH : 2Mg + O2 -> 2MgO
\(m_{O_2}=0,2.32=6,4\left(g\right)\)
b) Theo ĐLBTKL
\(m_{Mg}+m_{O_2}=m_{MgO}\)
\(=>m_{Mg}=8-6,4=1,6\left(g\right)\)