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Phần trăm khối lượng của nguyên tố S trong hợp chất nhôm sunfat là:
\(\%m_S=\dfrac{M_S.3}{M_{Al_2\left(SO_4\right)_3}}.100\%=\dfrac{32.3}{27.2+32.3+16.4.3}.100\%=\dfrac{16}{57}.100\%\approx28\%\)
\(a,n_{\left(NH_4\right)_3PO_4}=0,6\left(mol\right)\\ \Rightarrow n_N=0,6.3=1,8\left(mol\right)\Rightarrow m_N=1,8.14=25,2\left(g\right)\\ n_H=4.3.0,6=7,2\left(mol\right)\Rightarrow m_H=7,2.1=7,2\left(g\right)\\ n_P=n_{hc}=0,6\left(mol\right)\Rightarrow m_P=0,6.31=18,6\left(g\right)\\ n_O=4.0,6=2,4\left(mol\right)\Rightarrow m_O=2,4.16=38,4\left(g\right)\)
\(b,n_S=\dfrac{6,4}{32}=0,2\left(mol\right)\Rightarrow n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}.0,2=\dfrac{1}{15}\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=342.\dfrac{1}{15}=22,8\left(g\right)\\ c,n_{Al_2\left(SO_4\right)_3}=\dfrac{20,52}{342}=0,06\left(mol\right)\\ n_O=4.3.0,06=0,72\left(mol\right)\\ \Rightarrow n_{CO_2}=\dfrac{0,72}{2}=0,36\left(mol\right)\Rightarrow V_{CO_2\left(đktc\right)}=0,36.22,4=8,064\left(l\right)\)
Ta có: \(n_{Al}=\dfrac{12,15}{27}=0,45\left(mol\right)\)
\(n_{CuSO_4}=\dfrac{54}{160}=0,3375\left(mol\right)\)
PT: \(2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\)
Xét tỉ lệ: \(\dfrac{0,45}{2}>\dfrac{0,3375}{3}\), ta được Al dư.
Theo PT: \(n_{Al\left(pư\right)}=\dfrac{2}{3}n_{CuSO_4}=0,225\left(mol\right)\)
\(\Rightarrow n_{Al\left(dư\right)}=0,225\left(mol\right)\Rightarrow m_{Al\left(dư\right)}=0,225.27=6,075\left(g\right)\)
Theo PT: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{CuSO_4}=0,1125\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,1125.342=38,475\left(g\right)\)
\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)
\(Fe_2O_3+3H_2\underrightarrow{^{^{t^0}}}2Fe+3H_2O\)
\(b.\)
\(n_{Al_2\left(SO_4\right)_3}=\dfrac{68.4}{342}=0.2\left(mol\right)\)
Số nguyên tử Al :
\(0.2\cdot2\cdot6\cdot10^{23}=2.4\cdot10^{23}\left(pt\right)\)
Số nguyên tử S :
\(0.2\cdot3\cdot6\cdot10^{23}=3.6\cdot10^{23}\left(pt\right)\)
Số nguyên tử O :
\(0.2\cdot12\cdot6\cdot10^{23}=14.4\cdot10^{23}\left(pt\right)\)
\(FeCl_3:Fe\left(III\right)\\ SO_3:S\left(VI\right)\\ Mg\left(OH\right)_2:Mg\left(II\right)\\ Al_2\left(SO_4\right)_3:Al\left(III\right)\)
Câu 2 :
nAl2(SO4)3 = 68.4/342 = 0.2 (mol)
nS = 0.2*3 = 0.6 (mol)
Số nguyên tử S :
0.6 * 6 * 10^23 = 3.6 * 10^23 ( nguyên tử)
Câu 1:
n(XO,XO2)=11,2/22,4=0,5(mol)
=> M(hh)= 19,8/0,5= 39,6(g/mol)
Ta có:
\(M_{hh}=39,6\\ \Leftrightarrow\dfrac{2.\left(M_X+16\right)+3.\left(M_X+32\right)}{2+3}=39,6\\ \Leftrightarrow M_X=14\left(\dfrac{g}{mol}\right)\)
Vậy: X là Nitơ (N=14)
Chúc em học tốt!
\(n_{Al}=\dfrac{2,7}{27}=0,1mol\\ a.2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,1 0,15 0,05 0,15
\(b.V_{H_2}=0,15.24,79=3,7185l\\ c.m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1g\\ d.C_{M_{H_2SO_4}}=\dfrac{0,15}{0,4}=0,375M\)
\(\left\{{}\begin{matrix}\%Al=\dfrac{27.2}{342}.100\%=15,79\%\\\%S=\dfrac{32.3}{342}.100\%=28,07\%\\\%O=\dfrac{16.12}{342}.100\%=56,14\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%Fe=\dfrac{56.3}{232}.100\%=72,414\%\\\%O=\dfrac{4.16}{232}.100\%=27,586\%\end{matrix}\right.\)
a)
Giả sử có 100 gam hỗn hợp
\(m_O=\dfrac{25.100}{100}=25\left(g\right)\)
=> \(n_O=\dfrac{25}{16}=1,5625\left(mol\right)\)
Mà nO = 4.nS
=> \(n_S=\dfrac{1,5625}{4}=\dfrac{25}{64}\left(mol\right)\)
\(\%m_S=\dfrac{\dfrac{25}{64}.32}{100}.100\%=12,5\%\)
b) Đề bài cho rồi mà bn :)
c)
C1: %mkim loại = \(100\%-12,5\%-25\%=62,5\%\)
=> mkim loại = \(\dfrac{64.62,5}{100}=40\left(g\right)\)
C2:
\(m_S=\dfrac{64.12,5}{100}=8\left(g\right)\)
\(m_O=\dfrac{64.25}{100}=16\left(g\right)\)
=> mkim loại = 64 - 8 - 16 = 40 (g)
Ta có: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{68,4}{342}=0,2\left(mol\right)\)
\(\Rightarrow n_S=3n_{Al_2\left(SO_4\right)_3}=0,6\left(mol\right)\)
\(\Rightarrow m_S=0,6.32=19,2\left(g\right)\)
\(n_{Al_2\left(SO_4\right)_3}=\dfrac{68,4}{342}=0,2mol\\ m_S=32.3.0,2=19,2g\)