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a. Ag không phản ứng nên ta có PTHH: \(2Mg+O_2\rightarrow^{t^o}2MgO\)
\(\rightarrow m_{O_2}=m_{hh}-m_{\mu\text{ối}}=18,8-15,6=3,2g\)
\(\rightarrow n_{O_2}=\frac{3,2}{32}=0,1mol\)
b. \(\rightarrow V_{O_2}=n.22,4=22,4.0,1=2,24l\)
\(\rightarrow V_{kk}=4,48.5=11,2l\)
c. Có \(n_{Mg}=2n_{O_2}=0,2l\)
\(\rightarrow m_{Mg}=0,2.24=4,8g\)
\(\rightarrow\%m_{Mg}=\frac{4,8.100}{15,6}\approx30,77\%\)
\(\rightarrow\%m_{Ag}=100\%-30,77\%=69,23\%\)
1)
Gọi số mol KMnO4, KClO3 là a, b (mol)
=> 158a + 122,5b = 308,2 (1)
PTHH: 2KMnO4 --to--> K2MnO4 + MnO2 + O2
a-------------------------------->0,5a
2KClO3 --to--> 2KCl + 3O2
b------------------>1,5b
=> mO2 = (0,5a + 1,5b).32 = 16a + 48b (g)
mD = 308,2 - 16a - 48b(g)
\(m_{Mn}=\dfrac{\left(308,2-16a-48b\right).10,69}{100}=32,94658-1,7104a-5,1312b\left(g\right)\)
=> \(n_{Mn}=\dfrac{32,94658-1,7104a-5,1312b}{55}=0,6-\dfrac{1069}{34375}a-\dfrac{3207}{34375}\left(mol\right)\)
Mà \(n_{Mn}=n_{KMnO_4}=a\left(mol\right)\)
=> \(\dfrac{35444}{34375}a+\dfrac{3207}{34375}b=0,6\) (2)
(1)(2) => a = 0,4 (mol); b = 2 (mol)
=> \(\left\{{}\begin{matrix}\%m_{KMnO_4}=\dfrac{0,4.158}{308,2}.100\%=20,506\%\\\%m_{KClO_3}=\dfrac{2.122,5}{308,2}.100\%=79,494\%\end{matrix}\right.\)
2)
Giả sử nung 100 (g) đá vôi
=> \(m_{CaCO_3\left(bđ\right)}=\dfrac{80.100}{100}=80\left(g\right)\)
\(m_{rắn.sau.pư}=\dfrac{100.73,6}{100}=73,6\left(g\right)\)
=> mCO2 = 100 - 73,6 = 26,4 (g)
\(n_{CO_2}=\dfrac{26,4}{44}=0,6\left(mol\right)\)
PTHH: CaCO3 --to--> CaO + CO2
0,6<----------------0,6
=> mCaCO3(pư) = 0,6.100 = 60 (g)
\(H\%=\dfrac{60}{80}.100\%=75\%\)
a, PT: \(C+O_2\underrightarrow{t^o}CO_2\)
\(S+O_2\underrightarrow{t^o}SO_2\)
b, Giả sử: \(\left\{{}\begin{matrix}n_C=x\left(mol\right)\\n_S=y\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow12x+32y=5,6\left(1\right)\)
Ta có: \(n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\)
Theo PT: \(\Sigma n_{O_2}=n_C+n_S=x+y\left(mol\right)\)
\(\Rightarrow x+y=0,3\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_C=0,2.12=2,4\left(g\right)\\m_S=0,1.32=3,2\left(g\right)\end{matrix}\right.\)
c, Ta có: \(\left\{{}\begin{matrix}\%m_C=\dfrac{2,4}{5,6}.100\%\approx42,9\%\\\%m_S\approx57,1\%\end{matrix}\right.\)
d, Phần này đề yêu cầu tính theo khối lượng mol hả bạn?
Gọi số mol KClO3, KMnO4 trong mỗi phần là a, b (mol)
Phần 1:
\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
mY = 122,5a + 158b - 0,1.32 = 122,5a + 158b - 3,2 (g)
Bảo toàn O: \(n_{O\left(Y\right)}=3a+4b-0,2\left(mol\right)\)
\(\%O=\dfrac{16\left(3a+4b-0,2\right)}{122,5a+158b-3,2}.100\%=34,5\%\)
=> 5,7375a + 9,49b = 2,096 (1)
Phần 2:
PTHH: 2KClO3 --to--> 2KCl + 3O2
a----------->a
2KMnO4 --to--> K2MnO4 + MnO2 + O2
b------------>0,5b------>0,5b
=> 74,5a + 142b = 29,1 (2)
(1)(2) => a = 0,2 (mol); b = 0,1 (mol)
=> \(\left\{{}\begin{matrix}\%m_{KClO_3}=\dfrac{0,2.122,5}{0,2.122,5+0,1.158}.100\%=60,8\%\\\%m_{KMnO_4}=\dfrac{0,1.158}{0,2.122,5+0,1.158}.100\%=39,2\%\end{matrix}\right.\)
\(n_{Al}=\dfrac{1,728}{27}=0,064\left(mol\right)\)
PTHH: 4Al + 3O2 --to--> 2Al2O3
____0,064->0,048
=> mO2 = 0,048.32 = 1,536 (g)
\(m_B=\dfrac{0,894.100}{8,127}=11\left(g\right)\)
Theo ĐLBTKL: mA = mB + mO2
=> mA = 11 + 1,536 = 12,536 (g)
b. Gọi x,y lần lượt là số mol của C,S (x,y>0)
theo pương trình ta có:
x+y=\(\frac{9,6}{32}\)=0,3
12x+32y=5,6
=>\(\left\{\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}m_C=0,2.12=2,4g\\m_S=5,6-2,4=3,2g\end{matrix}\right.\)
c. %m\(_C\)=\(\frac{2,4}{5,6}.100\%=42,46\%\)
=>%m\(_S\)=100%-42,86%=57,14%
a. C+O\(_2\)\(\rightarrow\)CO\(_2\)
S+O\(_2\)\(\rightarrow\)SO\(_2\)
a, \(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)
\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)gdf
b, Gọi số mol KClO3 và KMnO4 lần lượt là x,y ( mol ) ( x,y > 0 )
\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)
x(mol)......................\(\dfrac{3}{2}x\left(mol\right)\)
\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)
y(mol)..............................................\(\dfrac{1}{2}y\left(mol\right)\)
\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
Tổng số mol O2 : \(\dfrac{3}{2}x+\dfrac{1}{2}y=0,05\left(1\right)\)
\(m_{KClO_3}=n.M=122,5x\left(g\right)\)
\(m_{KMnO_4}=n.M=158y\left(g\right)\)
\(\Rightarrow122,5x+158y=8,77\left(2\right)\)
Từ (1)(2) ,có :\(\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{2}y=0,05\\122,5x+158y=8,77\end{matrix}\right.\) ( bấm máy tính là ra )
\(\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,04\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{KClO_3}=122,5.x=122,5.0,02=2,45\left(g\right)\\m_{KMnO_4}=158y=158.0,04=6,32\left(g\right)\end{matrix}\right.\)
\(\%m_{KClO_3}=\dfrac{2,45}{8,77}.100\%=27\%\)
\(\%m_{KMnO_4}=\dfrac{6,32}{8,77}.100\%=73\%\)
nO2=1,12/22,4=0,05(mol)
2KClO3--->2KCl+3O2
x_______________3/2x
2KMnO4--->K2MnO4+MnO2+O2
y__________________________1/2y
Hệ pt:
\(\left\{{}\begin{matrix}122,5x+158y=8,77\\1,5x+0,5y=0,05\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,04\end{matrix}\right.\)
=>mKClO3=0,02.122,5=2,45(g)
=>%mKClO3=2,45/8,77.100%~27,9%
=>%mKMnO4=100%-27,9%=72,1%