500ml = 0,5l

\(n_{NaOH}=1.0,5=0,5\left(mol\right)\)

Pt : \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O|\)

            2                1                1              2

          0,5             0,25

\(n_{H2SO4}=\dfrac{0,5.1}{2}=0,25\left(mol\right)\)

\(V_{ddH2SO4}=\dfrac{0,25}{2}=0,125\left(l\right)\)

 Chúc bạn học tốt

2NaOH+H2SO4 -> Na2SO4+2H2O

n(NaOH)=0,5.1=0,5(mol)

Theo PTHH: n(H2SO4)=n(NaOH)/2

=> n(H2SO4)=0,25(mol)

=> V(H2SO4)=0,25/2=0,125l

Bài 2 : 

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

PTHH :

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

0,1           0,3                 0,1             0,3

\(m_{Fe_2\left(SO_4\right)_3}=0,1.400=40\left(g\right)\)

\(b,V_{ddH_2SO_4}=\dfrac{0,3}{2}=0,15\left(l\right)\)

\(c,C_{M\left(Fe_2\left(SO_4\right)_3\right)}=\dfrac{0,1}{0,15}=\dfrac{2}{3}\left(M\right)\)

Bài 3 :

\(n_{Mg}=\dfrac{4.8}{24}=0,2\left(mol\right)\)

PTHH :

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2O\)

0,2       0,2              0,2          0,2

\(m_{MgSO_4}=0,2.120=24\left(g\right)\)

\(V_{H_2}=0,2.24,79=4,958\left(l\right)\)

\(c,C_{M\left(H_2SO_4\right)}=\dfrac{0,2}{0,2}=1\left(M\right)\)

\(d,C_{M\left(MgSO_4\right)}=\dfrac{0,2}{0,2}=1\left(M\right)\)

Bài 4 :

\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

PTHH :

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

0,3       0,45              0,15          0,45

\(V_{H_2}=0,45.24,79=11,1555\left(l\right)\)

\(m_{H_2SO_4}=0,45.98=44,1\left(g\right)\)

\(C\%_{H_2SO_4}=\dfrac{44,1}{300}.100\%=14,7\%\)

\(m_{Al_2\left(SO_4\right)_3}=0,15.342=51,3\left(g\right)\)

\(m_{dd}=8,1+300-\left(0,45.2\right)=307,2\left(g\right)\)

\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{51,3}{307,2}.100\%\approx16,7\%\)

Bài 5 :

\(n_{H_2}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

PTHH:

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

0,2        0,4        0,2       0,2

\(m_{Fe}=0,2.56=11,2\left(g\right)\)

\(C_{M\left(HCl\right)}=\dfrac{0,4}{0,2}=2\left(M\right)\)

\(C_{M\left(FeCl_2\right)}=\dfrac{0,2}{0,2}=1\left(M\right)\)

1:

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

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

______0,2------>0,2------------------->0,2_____(mol)

=> \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b) \(V_{ddH_2SO_4}=\dfrac{0,2}{1}=0,2\left(l\right)\)

2:

a) 

\(n_{HCl}=2.0,2=0,4\left(mol\right)\)

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

______0,2<------0,4------------------>0,2______(mol)

=> \(m_{Mg}=0,2.24=4,8\left(g\right)\)

b) \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

cảm ơn bạn nhiều nha

nNaOH=0.01(mol)

2NaOH+SO2->Na2SO3+H2O

0.01 0.005

V=0.112(l)

2)nHCl=0.5(mol)

MgO+2HCl->MgCl2+H2O

x 2x

Fe2O3+6HCl->2FeCl3+3H2O

y 6y

Theo bài ra:40x+160y=12

2x+6y=0.5

x=0.1(mol) mMgO=4(g)

y=0.05(mol) mFe2O3=8(g)

bn giải câu 1 hay câu 2 z

\(n_{CaCO3}=\dfrac{30}{100}=0,3\left(mol\right)\)

\(n_{NaOH}=2.0,25=0,5\left(mol\right)\)

a) \(CaCO3+2HCl-->CaCl2+CO2+H2O\)

......0,3....................................................0,3............

\(k=\dfrac{0,5}{0,3}=1,67\)

\(CO_2+2NaOH-->Na_2CO_3+H_2O\)

..x/2.............x

\(CO_2+NaOH-->NaHCO3\)

y...............y.....

bn tự giải hệ ra nhé. đến đây thì dễ r

Bài 1

\(a,n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\ CuO+2HCl\xrightarrow[]{}CuCl_2+H_2O\\ n_{CuCl_2}=n_{CuO}=0,2mol\\ m_{CuCl_2}=0,2.135=27\left(g\right)\\ b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ C_{MHCl}=\dfrac{0,4}{0,5}=0,8\left(M\right)\)

Bài 5

\(a,n_{NaOH}=0,2.1=0,2\left(mol\right)\\ 2NaOH+H_2SO_4\xrightarrow[]{}Na_2SO_4+2H_2O\\ n_{H_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MH_2SO_4}=\dfrac{0,1}{0,4}=0,25\left(M\right)\\ b,n_{Na_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MNa_2SO_4}=\dfrac{0,1}{0,2+0,4}=\dfrac{1}{6}\left(M\right)\\ c,m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)