Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt :
nCO = a (mol)
nCO2 = b (mol)
=> a + b = 0.4 (1)
mX = 28a + 44b = 40(a+b) (g)
=> 12a -4b = 0 (2)
(1) , (2):
a = 0.1
b = 0.3
mC = ( 0.1 + 0.3) * 12 = 4.8(g)
$\%m_{O_2(X)}=\dfrac{1,6}{1,6+4,4}.100\%=26,67\%$
$n_{CO_2}=\dfrac{4,4}{44}=0,1(mol);n_{O_2}=\dfrac{1,6}{16}=0,05(mol)$
$\Rightarrow \%V_{O_2(X)}=\dfrac{0,05}{0,05+0,1}.100\%=33,33\%$
$C+O_2\xrightarrow{t^o}CO_2$
Theo PT: $n_C=n_{O_2(p/ứ)}=n_{CO_2}=0,1(mol)$
$\Rightarrow n_{O_2(dùng)}=0,1+0,05=0,15(mol)$
$m_C=0,1.12=1,2(g);V_{O_2(dùng)}=0,15.22,4=3,36(lít)$
$\to m=1,2;V=3,36$
Có \(\left\{{}\begin{matrix}n_{H_2}+n_{C_2H_2}=\dfrac{17,92}{22,4}=0,8\\\dfrac{2.n_{H_2}+26.n_{C_2H_2}}{n_{H_2}+n_{C_2H_2}}=0,5.28=14\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_{H_2}=0,4\left(mol\right)\\n_{C_2H_2}=0,4\left(mol\right)\end{matrix}\right.\)
\(n_{O_2}=\dfrac{51,2}{32}=1,6\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
0,4-->0,2
2C2H2 + 5O2 --to--> 4CO2 + 2H2O
0,4----->1------------>0,8
=> Y chứa \(\left\{{}\begin{matrix}CO_2:0,8\left(mol\right)\\O_{2\left(dư\right)}:0,4\left(mol\right)\end{matrix}\right.\)
=> \(\overline{M}_Y=\dfrac{0,8.44+0,4.32}{0,8+0,4}=40\left(g/mol\right)\)
\(\overline{M}_X=14\left(g/mol\right)\)
=> \(d_{X/Y}=\dfrac{14}{40}=0,35\)
Z gồm CO2 và O2 dư
$C + O_2 \xrightarrow{t^o} CO_2$
$n_{CO_2} =n_{O_2\ pư} = n_C = \dfrac{1,128}{12} = 0,094(mol)$
Gọi $n_{O_2} = 2a \to n_{không\ khí} = 3a(mol)$
Trong Y :
$n_{O_2} = 2a + 3a.20\% = 2,6a(mol)$
$n_{N_2} = 3a.80\% = 2,4a(mol)$
Trong Z :
$n_{CO_2} = 0,094(mol)$
$n_{N_2} = 2,4a(mol)$
$n_{O_2\ dư} = n_{O_2} - n_{O_2\ pư} = 2,6a - 0,094(mol)$
m CO2 =0,094.44 = 4,136(gam)
=> m Z = 4,136 : 27,5% = 15,04(gam)
SUy ra :
4,136 + 2,4a.28 + (2,6a - 0,094).32 = 15,04
=> a = 0,0925
=> n O2 = 0,0925.2 = 0,185(mol)
m X = 43,5 : 46,4% = 93,75(gam)
Bảo toàn khối lượng : m = 93,75 + 0,185.32 = 99,67(gam)
a)
dA/O\(_2\) = \(\dfrac{M_A}{32}\) = 1,25 \(\Rightarrow\) MA = 32 . 1,25 = 40
PTPƯ: C + O2 -----> CO2
C + CO2 -----> 2CO
Trường hợp 1 (Oxi dư)
Ta có: MA = \(\dfrac{44x+\left(1-x\right).32}{1}\) = 40 \(\Rightarrow\) x = \(\dfrac{2}{3}\)
Vậy %VCO\(_2\) = \(\dfrac{2}{3}\) . 100 = 66,67%
%VO\(_2\) = 33,33%
Trường hợp 2 (Oxi thiếu)
MA = \(\dfrac{44x+\left(1-x\right).28}{1}\) = 40 \(\Rightarrow\) x = 0,75
Vậy % VCO\(_2\) = \(\dfrac{a}{a+b}\) . 100 = \(\dfrac{3b}{4b}\) . 100 = 75%
%VCO = 25%
b)
CO2 + CA(OH)2 -----> CaOH3 \(\downarrow\) + H2O
0,06 \(\leftarrow\) 0,06 = \(\dfrac{6}{100}\)
Trường hợp 1 (nCO\(_2\) = 0,06 mol \(\Rightarrow\) nO\(_2\) dư = 0,03 mol)
Vậy mc = 0,06.12 = 0,75 (g)
VO\(_2\) = (0,06 + 0,03) . 22,4 = 2,016 (l)
Trường hợp 2 (nCO\(_2\) = 0,06 mol, nCO = \(\dfrac{1}{3}\) nCO\(_2\) = 0,02 mol)
\(\Rightarrow\) nC = nCO\(_2\) + nCO = 0,06 + 0,02 = 0,08 (mol)
\(\Rightarrow\) mC = 0,08 . 12 = 0,96 (g)
nO\(_2\) = nCO\(_2\) + \(\dfrac{1}{2}\) nCO = 0,06 + 0,01 = 0,07 (mol)
VO\(_2\) = 0,07.22,4 = 1,568 (l)
cho mình hỏi tại sao ở câu b th 1 no2 dư = 0,03 với còn th 2 thì nco = 1/3 nco2
PTHH: \(2Mg+O_2\underrightarrow{t^o}2MgO\) (1)
\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\) (2)
a) Gọi số mol của Mg là a (mol) \(\Rightarrow n_{Al}=\dfrac{2}{3}a\left(mol\right)\)
\(\Rightarrow24a+27\cdot\dfrac{2}{3}a=6,3\) \(\Rightarrow a=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{MgO}=0,15\left(mol\right)\\n_{Al_2O_3}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgO}=0,15\cdot40=6\left(g\right)\\m_{Al_2O_3}=0,05\cdot102=5,1\left(g\right)\end{matrix}\right.\)
b) Theo các PTHH: \(\left\{{}\begin{matrix}n_{O_2\left(1\right)}=0,075\left(mol\right)\\n_{O_2\left(2\right)}=0,075\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\Sigma n_{O_2}=0,15\left(mol\right)\) \(\Rightarrow V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\)
a) PTHH: \(2CO+O_2\underrightarrow{t^o}2CO_2\) (1)
\(4H_2+O_2\underrightarrow{t^o}2H_2O\) (2)
b) Ta có: \(\left\{{}\begin{matrix}\Sigma n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\\n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{O_2\left(1\right)}=0,1mol\\n_{O_2\left(2\right)}=0,2mol\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CO}=0,1\cdot28=2,8\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{CO}=\dfrac{2,8}{2,8+0,4}\cdot100\%=87,5\%\\\%m_{H_2}=12,5\%\end{matrix}\right.\)
c) PTHH: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
Theo PTHH: \(n_{KMnO_4}=2n_{O_2}=0,6mol\)
\(\Rightarrow m_{KMnO_4}=0,6\cdot158=94,8\left(g\right)\)
a) Gọi số mol H2, CH4 là a, b
=> \(a+b=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(M_X=\dfrac{2a+16b}{a+b}=0,325.32=10,4\)
=> a = 0,2 ; b = 0,3
=> \(\left\{{}\begin{matrix}\%V_{H_2}=\dfrac{0,2}{0,5}.100\%=40\%\\\%V_{CH_4}=\dfrac{0,3}{0,5}.100\%=60\%\end{matrix}\right.\)
b) \(n_{O_2}=\dfrac{32}{32}=1\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,3--->0,6------->0,3
2H2 + O2 --to--> 2H2O
0,2-->0,1
=> \(\left\{{}\begin{matrix}V_{CO_2}=0,3.22,4=6,72\left(l\right)\\V_{O_2\left(dư\right)}=\left(1-0,6-0,1\right).22,4=6,72\left(l\right)\end{matrix}\right.\)
Gọi số mol R, O2 là a, b (mol)
=> m1 = a.MR; m2 = 32b
PTHH: R + O2 --to--> RO2
a->a-------->a
=> hh khí gồm \(\left\{{}\begin{matrix}RO_2:a\left(mol\right)\\O_{2\left(dư\right)}:b-a\left(mol\right)\end{matrix}\right.\)
Xét \(\overline{M}=\dfrac{a\left(M_R+32\right)+32\left(b-a\right)}{a+\left(b-a\right)}=25,6.2=51,2\)
=> a.MR = 19,2b
Xét \(\dfrac{m_1}{m_2}=\dfrac{a.M_R}{32b}=\dfrac{19,2b}{32b}=\dfrac{3}{5}\)
hi lo ae