Phần trăm khối lượng của nguyên tố S trong hợp chất nhôm sunfat là:

\(\%m_S=\dfrac{M_S.3}{M_{Al_2\left(SO_4\right)_3}}.100\%=\dfrac{32.3}{27.2+32.3+16.4.3}.100\%=\dfrac{16}{57}.100\%\approx28\%\)

\(\%S=\dfrac{96}{342}=28,07\%\)

\(a,n_{\left(NH_4\right)_3PO_4}=0,6\left(mol\right)\\ \Rightarrow n_N=0,6.3=1,8\left(mol\right)\Rightarrow m_N=1,8.14=25,2\left(g\right)\\ n_H=4.3.0,6=7,2\left(mol\right)\Rightarrow m_H=7,2.1=7,2\left(g\right)\\ n_P=n_{hc}=0,6\left(mol\right)\Rightarrow m_P=0,6.31=18,6\left(g\right)\\ n_O=4.0,6=2,4\left(mol\right)\Rightarrow m_O=2,4.16=38,4\left(g\right)\)

\(b,n_S=\dfrac{6,4}{32}=0,2\left(mol\right)\Rightarrow n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}.0,2=\dfrac{1}{15}\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=342.\dfrac{1}{15}=22,8\left(g\right)\\ c,n_{Al_2\left(SO_4\right)_3}=\dfrac{20,52}{342}=0,06\left(mol\right)\\ n_O=4.3.0,06=0,72\left(mol\right)\\ \Rightarrow n_{CO_2}=\dfrac{0,72}{2}=0,36\left(mol\right)\Rightarrow V_{CO_2\left(đktc\right)}=0,36.22,4=8,064\left(l\right)\)

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: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{CuSO_4}=0,1125\left(mol\right)\)

\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,1125.342=38,475\left(g\right)\)

a, CaO

b, Al2(SO4)3

\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)

\(Fe_2O_3+3H_2\underrightarrow{^{^{t^0}}}2Fe+3H_2O\)

\(b.\)

\(n_{Al_2\left(SO_4\right)_3}=\dfrac{68.4}{342}=0.2\left(mol\right)\)

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

\(b.V_{H_2}=0,15.24,79=3,7185l\\ c.m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1g\\ d.C_{M_{H_2SO_4}}=\dfrac{0,15}{0,4}=0,375M\)

D