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

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

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

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

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

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

1. 

a, K₂O  +  H₂O --> 2KOH (1:1:2)

b, 2Cu + O₂ --> 2CuO (2:1:2)

c, Al₂(SO₄)₃  + 3BaCl₂ -->  3BaSO₄  +  2AlCl₃ (1:3:3:2)

Bài 2:

\(a,n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{H_2O}=\dfrac{5,4}{18}=0,3\left(mol\right)\\ n_{CO_2}=\dfrac{13,2}{44}=0,3\left(mol\right)\\ b,n_{NaCl}=\dfrac{0,6.10^{23}}{6.10^{23}}=0,1\left(mol\right)\\ n_{Ca}=\dfrac{1,2.10^{23}}{6.10^{23}}=0,2\left(mol\right)\)

Bài 3:

\(a,m_{H_2SO_4}=0,5.98=49\left(g\right)\\ m_{NaOH}=0,2.40=8\left(g\right)\\ m_{Ag}=108.0,1=10,8\left(g\right)\\ b,n_{SO_2}=\dfrac{15.10^{23}}{6.10^{23}}=2,5\left(mol\right)\\ m_{SO_2}=64.2,5=160\left(g\right)\)

Có \(PTK_{Al_2\left(SO_4\right)_x}=342đvC\)

\(\rightarrow27.2+\left(32+16.4\right).x=342\)

\(\rightarrow54+96x=342\)

\(\rightarrow x=3\)

2Al + 3 CuSO4 -> 3Cu + Al2(SO4)3

Số nguyên tử Cu tạo thành khi có 150 nguyên tử Al phản ứng:

(150 x 3)/2= 225(nguyên tử)