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a: \(C+O_2\rightarrow CO_2\)(ĐK: t độ)
x x x
\(S+O_2\rightarrow SO_2\)(ĐK: t độ)
y y y
b: \(n_{O_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)
Theo đề, ta có hệ:
12x+32y=10 và x+y=0,5
=>x=0,3 và y=0,2
\(m_C=0.3\cdot12=3.6\left(g\right)\)
\(m_S=0.2\cdot32=6.4\left(g\right)\)
c: \(n_{CO_2}=n_C=0.3\left(mol\right)\)
\(n_{SO_2}=n_S=0.2\left(mol\right)\)
\(V_{khí}=22.4\left(0.3+0.2\right)=11.2\left(lít\right)\)
\(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
PTHH :
\(C+O_2\rightarrow\left(t^o\right)CO_2\)
x x x
\(S+O_2\rightarrow\left(t^o\right)SO_2\)
y y y
Gọi n C = x
n S = y (mol)
Ta có hệ PT :
\(\left\{{}\begin{matrix}12x+32y=10\\x+y=0,5\end{matrix}\right.\)
\(\rightarrow x=0,3;y=0,2\)
\(m_C=0,3.12=3,6\left(g\right)\)
\(m_S=0,2.32=6,4\left(g\right)\)
\(c,V_{hhk}=\left(0,3+0,2\right).22,4=11,2\left(l\right)\)
Đặt \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
Theo đề: \(m_{hh}=36\left(g\right)\)
\(\Rightarrow m_{Mg}+m_{Fe}=36\\ \Rightarrow24x+56y=36\left(1\right)\)
\(PTHH:2Mg+O_2\underrightarrow{t^o}2MgO\\ \left(mol\right)....x\rightarrow...0,5x.....x\\ PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ \left(mol\right)....y\rightarrow...\dfrac{2}{3}y....\dfrac{1}{3}y\)
Theo đề: \(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
\(\Rightarrow0,5x+\dfrac{2}{3}y=0,6\left(2\right)\)
\(\xrightarrow[\left(2\right)]{\left(1\right)}\left\{{}\begin{matrix}24x+56y=36\\0,5x+\dfrac{2}{3}y=0,6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0,8\\y=0,3\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,8.24=19,2\left(g\right)\\m_{Fe}=0,3.56=16,8\left(g\right)\end{matrix}\right.\\ m_r=m_{MgO}+m_{Fe_3O_4}=0,8.40+\dfrac{1}{3}.0,3.232=55,2\left(g\right)\)
PTHH: C+O2→CO20,3mol:0,3mol→0,3molC+O2→CO20,3mol:0,3mol→0,3mol
S+O2→SO20,2mol:0,2mol→0,2molS+O2→SO20,2mol:0,2mol→0,2mol
mC=36%10100%=3,6(g)⇔nC=3,612=0,3(mol)mC=36%10100%=3,6(g)⇔nC=3,612=0,3(mol)
mS=10−3,6=6,4(g)⇔nS=6,432=0,2(mol)mS=10−3,6=6,4(g)⇔nS=6,432=0,2(mol)
VO2=(0,3+0,2)22,4=11,2(l)VO2=(0,3+0,2)22,4=11,2(l)
mhh=mCO2+mSO2=0,3.44+0,2.64=26(g)
Đặt \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
Theo đề: \(m_{hh}=39\left(g\right)\)
\(\Rightarrow m_{Al}+m_{Fe}=39\\ \Rightarrow27x+56y=39\left(1\right)\)
\(PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\\ \left(mol\right)....x\rightarrow..0.75x....0,5x\\ PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ \left(mol\right)....y\rightarrow..\dfrac{2}{3}y.....\dfrac{1}{3}y\)
Theo đề: \(n_{O_2}=\dfrac{V}{22,4}=\dfrac{12,32}{22,4}=0,55\left(mol\right)\)
\(\Rightarrow0,75x+\dfrac{2}{3}y=0,55\left(2\right)\)
\(\xrightarrow[\left(1\right)]{\left(2\right)}\left\{{}\begin{matrix}27x+56y=39\\0,75x+\dfrac{2}{3}y=0,55\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,6\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Fe}=0,6.56=33,6\left(g\right)\end{matrix}\right.\\ m_r=m_{Al_2O_3}+m_{Fe_3O_4}=0,5.0,2.102+\dfrac{1}{3}.0,6.232=56,6\left(g\right)\)
\(m_C=10.36\%=3,6\left(g\right)\Rightarrow n_C=0,3\left(mol\right)\)
\(m_S=10-3,6=6,4\left(g\right)\Rightarrow n_S=0,2\left(mol\right)\)
\(C+O_2-^{t^o}\rightarrow CO_2\)
\(S+O_2-^{t^o}\rightarrow SO_2\)
\(n_{O_2}=0,3+0,2=0,5\left(mol\right)\)
Vì oxi chiếm 20% thể tích không khí
=> \(V_{kk}=\dfrac{0,5.22,4}{20\%}=56\left(lít\right)\)
\(n_{hh}=n_{CO_2}+n_{SO_2}=0,3+0,2=0,5\left(mol\right)\)
=> \(V_{hh}=0,5.22,4=11,2\left(lít\right)\)
PTHH: C+O2→CO20,3mol:0,3mol→0,3molC+O2→CO20,3mol:0,3mol→0,3mol
S+O2→SO20,2mol:0,2mol→0,2molS+O2→SO20,2mol:0,2mol→0,2mol
mC=36%10100%=3,6(g)⇔nC=3,612=0,3(mol)mC=36%10100%=3,6(g)⇔nC=3,612=0,3(mol)
mS=10−3,6=6,4(g)⇔nS=6,432=0,2(mol)mS=10−3,6=6,4(g)⇔nS=6,432=0,2(mol)
VO2=(0,3+0,2)22,4=11,2(l)VO2=(0,3+0,2)22,4=11,2(l)
mhh=mCO2+mSO2=0,3.44+0,2.64=26(g)
Gọi x là số mol của CH 4 => V CH 4 = n.22,4 = 22,4x
y là số mol của H 2 => V H 2 = 22,4y
V hh = V H 2 + V CH 4 => 22,4x + 22,4y = 11,2
n H 2 O = m/M = 16,2/18 = 0,9 mol
n H 2 O = 2x + y = 0,9
Từ (1) và (2), ta có hệ phương trình:
22,4x + 22,4y = 11,2
2x + y = 0,9
Giải hệ phương trình ta có: x = 0,4( mol); y= 0,1 mol
V CH 4 = 22,4x = 22,4x0,4 = 8,96l
% V CH 4 = 8,96/11,2 x 100% = 80%
% V H 2 = 100% - 80% = 20%
Gọi số mol CH4, C2H6 là a, b
=> a+b = \(\dfrac{5,6}{22,4}=0,25\)
Có \(\dfrac{16a+30b}{a+b}=0,6.29=17,4\)
=> a = 0,225; b = 0,025
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
____0,225->0,45------->0,225->0,45
2C2H6 + 7O2 --to--> 4CO2 + 6H2O
0,025->0,0875----->0,05-->0,075
=> VO2 = (0,45+0,0875).22,4 = 12,04 (l)
mCO2 = (0,225 + 0,05).44 = 12,1(g)
mH2O = (0,45+ 0,075).18 = 9,45(g)
\(CuO+CO\rightarrow Cu+CO_2\)
..x..........x.........................
\(PbO+CO\rightarrow Pb+CO_2\)
..y........y........................
- Theo bài ra ta có hệ : \(\left\{{}\begin{matrix}80x+223y=3,83\\x+y=0,03\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,01\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=1,6\\m_{PbO}=2,23\end{matrix}\right.\) ( g )
b, \(n_K=n_{CO_2}=x+y=0,03\left(mol\right)\)
\(\Rightarrow V=0,672\left(l\right)\)
c, \(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)
........................0,03........0,03.............
\(\Rightarrow m_{kt}=3\left(g\right)\)
Đặt \(\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{PbO}=y\left(mol\right)\end{matrix}\right.\)
\(m_{CuO}+m_{PbO}=3,83\\ \Rightarrow80x+223y=3,83\left(1\right)\)
\(PTHH:CuO+CO\underrightarrow{t^o}Cu+CO_2\uparrow\\ \left(mol\right)......x\rightarrow..x....x.....x\\ PTHH:PbO+CO\underrightarrow{t^o}Pb+CO_2\uparrow\\ \left(mol\right)......y\rightarrow..y....y.....y\\ n_{CO}=\dfrac{0,84}{28}=0,03\\ \Rightarrow x+y=0,03\left(2\right)\)
Từ (1) và (2) ta có hpt \(\left\{{}\begin{matrix}80x+223y=3,83\\x+y=0,03\end{matrix}\right.\)
Giải hpt ta được \(\left\{{}\begin{matrix}x=0,02\\y=0,01\end{matrix}\right.\)
\(a,\left\{{}\begin{matrix}m_{CuO}=80.0,02=1,6\left(g\right)\\m_{PbO}=3,83-1,6=2,23\left(g\right)\end{matrix}\right.\)
\(b,V_{CO_2}=\left(x+y\right).22,4=\left(0,02+0,01\right).22,4=0,672\left(l\right)\)
\(c,n_{CO_2}=x+y=0,02+0,01=0,03\left(mol\right)\\ PTHH:Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\\ \left(mol\right)................0,03\rightarrow0,03\\ m_{CaCO_3}=0,03.100=3\left(g\right)\)
a, PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
Giả sử: \(\left\{{}\begin{matrix}n_{C_2H_4}=x\left(mol\right)\\n_{C_2H_2}=y\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow x+y=\dfrac{1,344}{22,4}=0,06\left(1\right)\)
Ta có: \(n_{Br_2}=\dfrac{16}{160}=0,1\left(mol\right)\)
Theo PT: \(n_{Br_2}=n_{C_2H_4}+2n_{C_2H_2}=x+2y\left(mol\right)\)
⇒ x + 2y = 0,1 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,02\left(mol\right)\\y=0,04\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,02}{0,06}.100\%\approx33,33\%\\\%\text{ }V_{C_2H_2}\approx66,67\%\end{matrix}\right.\)
b, Ta có: 1/2 hỗn hợp khí gồm: 0,01 mol C2H4 và 0,02 mol C2H2.
PT: \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)
Theo PT: \(n_{CO_2}=2n_{C_2H_4}+2n_{C_2H_2}=0,06\left(mol\right)\)
\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)
Theo PT: \(n_{CaCO_3}=n_{CO_2}=0,06\left(mol\right)\)
\(\Rightarrow m_{cr}=m_{CaCO_3}=0,06.100=6\left(g\right)\)
Bạn tham khảo nhé!
PTHH: \(C+O_2\xrightarrow[]{t^o}CO_2\)
a___a______a (mol)
\(S+O_2\xrightarrow[]{t^o}SO_2\)
b___b_______b (mol)
Ta lập HPT: \(\left\{{}\begin{matrix}12a+32b=10\\a+b=\dfrac{11,2}{22,4}=0,5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,3\\b=0,2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_C=0,3\cdot12=3,6\left(g\right)\\m_S=6,4\left(g\right)\\V_{khí}=0,5\cdot22,4=11,2\left(l\right)\end{matrix}\right.\)