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

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

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!

a) 

\(n_{Al}=0,2.2=0,4\left(mol\right)\)

\(n_S=0,2.3=0,6\left(mol\right)\)

\(n_O=0,2.12=2,4\left(mol\right)\)

b)

\(n_N=0,5.2=1\left(mol\right)\)

\(n_H=0,5.4=2\left(mol\right)\)

\(n_O=0,5.3=1,5\left(mol\right)\)

Gọi số mol FeSO₄, Al₂(SO₄)₃ là a, b (mol)

n_Fe = a (mol)

n_Al = 2b (mol)

n_S = a + 3b (mol)

n_O = 4a + 12b (mol)

Có: \(\dfrac{n_O}{\Sigma n}=\dfrac{4a+12b}{a+2b+a+3b+4a+12b}=\dfrac{20}{29}\)

=> a = 2b

\(\left\{{}\begin{matrix}\%m_{FeSO_4}=\dfrac{152a}{152a+342b}.100\%=\dfrac{152.2b}{152.2b+342b}.100\%=47,059\%\\\%m_{Al_2\left(SO_4\right)_3}=100\%-47,059\%=52,941\%\end{matrix}\right.\)

a, \(n_{Al_2\left(SO_4\right)_3}=\dfrac{42,75}{342}=0,125\left(mol\right)\)

\(n_O=12n_{Al_2\left(SO_4\right)_3}=1,5\left(mol\right)\)

b, \(n_{O\left(Al_2O_3\right)}=3n_{Al_2O_3}=2n_{O\left(Al_2\left(SO_4\right)_3\right)}=3\left(mol\right)\)

\(\Rightarrow n_{Al_2O_3}=1\left(mol\right)\)

\(\Rightarrow m_{Al_2O_3}=102\left(g\right)\)

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

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

\(b.\)

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

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 : 

\(0.2\cdot12\cdot6\cdot10^{23}=14.4\cdot10^{23}\left(pt\right)\)

Gọi số nguyên tử của \(Al_2(SO_4)_3\) và \(K_2SO_4\) lần lượt là x và y.

Số nguyên tử của \(Al_2(SO_4)_3\) là \(2+3+4\cdot3=17x\)

Số nguyên tử của \(K_2SO_4\) là \(2+1+4=7y\)

Mà số nguyên tử \(O_2\) trong hỗn hợp là \(4\cdot3x+4y=12x+4y\)

Theo bài: \(n_{O_2}=\dfrac{20}{31}n_{hh}\)

\(\Rightarrow12x+4y=\dfrac{20}{31}\left(17x+7y\right)\)

\(\Rightarrow y=2x\)

Có \(\%m_{Al_2\left(SO_4\right)_3}=\dfrac{342x}{342x+174y}\cdot100\%=\dfrac{342x}{342x+174\cdot2x}\cdot100\%=49,56\%\)

\(\%m_{K_2SO_4}=100\%-49,56\%=50,44\%\)

\(\%m_{\dfrac{O}{Nhôm.sunfat}}=\dfrac{4.16.3}{342}.100\approx56,14\%\\ \%m_{\dfrac{O}{Kali.sunfat}}=\dfrac{4.16}{174}.100\approx36,78\%\\ Gọi:a=n_{Al_2\left(SO_4\right)_3};b=n_{K_2SO_4}\left(a,b>0\right)\\ \Rightarrow Vì:m_{\dfrac{O}{hh}}=\dfrac{20}{31}\\ \Leftrightarrow\dfrac{a.12+b.4}{17a+7b}.100\%=\dfrac{20}{31}\\ \Leftrightarrow32a=16b\\ \Leftrightarrow\dfrac{a}{b}=\dfrac{16}{32}=\dfrac{1}{2}\\ \Leftrightarrow b=2a\\ \%m_{\dfrac{K_2SO_4}{hh}}=\dfrac{174.2a}{174.2a+342.a}.100\%\approx50,435\%\\ \Rightarrow\%m_{\dfrac{Al_2\left(SO_4\right)_3}{hh}}\approx49,565\%\)

\(a) n_{Zn(NO_3)_2} = \dfrac{37,8}{189} = 0,2(mol)\\ n_{Zn} = 0,2\ mol \to m_{Zn} = 0,2.65 = 13\ gam\\ n_N = 0,2.2 = 0,4\ mol \to m_N = 0,4.14 = 5,6\ gam\\ m_O = 37,5 - 13 - 5,6 = 18,9(gam)\\ b)n_{Fe_3(PO_4)_2} = \dfrac{10,74}{358} = 0,03(moL)\\ n_{Fe} = 0,03.3 = 0,09 \to m_{Fe} = 0,09.56 = 5,04(gam)\\ n_P = 0,03.2 = 0,06 \to m_P = 0,06.31 = 1,86(gam)\\ m_O = 10,74 - 5,04 -1,86 = 3,84(gam)\\ c) n_{Al} = 0,2.2 = 0,4(mol\to m_{Al} = 0,4.27 = 10,8(gam)\\ n_S = 0,2.3 = 0,6 \to m_S = 0,6.32 = 19,2(gam)\\ n_O = 0,2.12 = 2,4 \to m_O = 2,4.16 = 38,4(gam)\)

\(d) n_{Zn(NO_3)_2} = \dfrac{6.10^{20}}{6.10^{23}} = 0,001(mol)\\ n_{Zn} = 0,001 \to m_{Zn} = 0,001.65 = 0,065(gam)\\ n_N = 0,001.2 = 0,002 \to m_N = 0,002.14 = 0,028(gam)\\ n_O = 0,001.6 = 0,006 \to m_O = 0,006.16= 0,096(gam)\)

Theo gt ta có: $n_{Zn(NO_3)_2}=0,2(mol);n_{Fe_3(PO_4)_2}=0,03(mol);n_{Zn(NO_3)_2}=1(mol)$

a, $m_{Zn}=13(g);m_{N}=5,6(g);m_{O}=19,2(g)$

b, $m_{Fe}=5,04(g);m_{P}=1,86(g)$;m_{O}=3,84(g)$

c, $m_{Al}=10,8(g);m_{S}=19,2(g);m_{O}=38,4(g)$

d, $m_{Zn}=65(g);m_{N}=28(g);m_{O}=96(g)$