\(m_{NaCl}=500.0,1\%=0,5\left(g\right);m_{H_2O}=500-0,5=499,5\left(g\right)\Rightarrow V_{H_2O}=\dfrac{499,5}{1}=499,5\left(l\right)\)

Pha chế: Lấy 0,5 gam tinh thể muối ăn NaCl và đổ 499,5ml H₂O

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ a,PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{HCl}=\dfrac{6}{2}.0,1=0,3\left(mol\right)\\ b,V=V_{ddHCl}=\dfrac{0,3}{3}=0,1\left(l\right)=100\left(ml\right)\)

Câu a thiếu đề em

7. \(H_2SO_4\): Axit sunfuric

\(Fe\left(OH\right)_2\) : Sắt(II) hiđroxit

\(NaCl\) : Natri clorua

\(P_2O_5\):  Điphotphopenta oxit

8. \(FeCl_3\): Sắt (III) clorua

\(Fe\left(OH\right)_3\): Sắt(III) hiđroxit

\(HCl\): Axit clohiđric

\(SO_3\): Lưu huỳnh trioxit

Điphotpho pentaoxit viết tách như vậy

Anh có làm rồi nha!

\(2Na+2H_2O\rightarrow2NaOH+H_2\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Na}=2.0,3=0,6\left(mol\right)\\ a,m_{Na}=0,6.23=13,8\left(g\right)\\ m_{Na_2O}=26,2-13,8=12,4\left(g\right)\\b, n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\\ n_{NaOH\left(tổng\right)}=n_{Na}+2.n_{Na_2O}=0,6+\dfrac{12,4}{62}=0,8\left(mol\right)\\ m_{c.tan}=m_{NaOH}=0,8.40=32\left(g\right)\\ c,m_{ddNaOH}=m_{hh}+m_{H_2O}-m_{H_2}=26,2+200-0,3.2=225,6\left(g\right)\\ C\%_{ddNaOH}=\dfrac{32}{225,6}.100\approx14,185\%\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH :

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

0,2     0,2 

\(V_{H_2SO_4}=\dfrac{0,2}{2}=0,1\left(l\right)\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\PTHH:Zn+H_2SO_4\xrightarrow[]{} ZnSO_4+H_2\\ \Rightarrow n_{Zn}=n_{H_2SO_4}=0,2\left(mol\right)\\ V_{H_2SO_4}=\dfrac{0,2}{2}=0,1\left(lít\right)\)

\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

0,6      0,4         0,2

\(n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2\left(mol\right)\)

\(a,m_{Fe}=0,6.56=33,6\left(g\right)\)

\(b,m_{O_2}=0,4.32=12,8\left(g\right)\)

Axit : 

H3PO4 : Axit photphoric 

HBr : Axit Bromhidric

Bazo : Mg(OH)2 : Magie hidroxit

Cu(OH)2 : Đồng (II) Hidroxit 

Muối :

NaSiO3 : Natri Silicat 

Fe2(SO4)3 : Sắt (III) Sunfat 

Oxit : N2O5 : Đi nito pentaoxit 

a, \(2Zn+O_2\underrightarrow{t^o}2ZnO\) - pư hóa hợp

b, \(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\) - pư phân hủy

c, \(CaO+H_2O\rightarrow Ca\left(OH\right)_2\) - pư hóa hợp

d, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\) - pư phân hủy