Question

Answer 1

Bài 2 :

\(n_{NaCl}=C_M.V=0,3\left(mol\right)\)

\(n_{AgNO3}=C_M.V=0,45\left(mol\right)\)

\(PTHH:AgNO_3+NaCl\rightarrow NaNO_3+AgCl\)

Từ PTHH : Thấy sau phản ứng AgNO3 dư ( 0,45 - 0,3 = 0,15 mol )

- Chắc là coi thể tích không đổi .

\(\Rightarrow C_{MAgNO3}=\dfrac{n}{V}=\dfrac{0,15}{0,15+0,15}=0,5M\)

Vậy ...

Answer 2

Bài 3 :

a, 4Na + O2 -> 2Na2O

Na2O + H2O -> 2NaOH

2NaOH + CO2 -> Na2CO3 + H2O

Na2CO3 + H2SO4 -> Na2SO4 + H2O + CO2

b, Mg + O2 -> MgO

MgO + H2SO4 -> MgSO4 + H2O

MgSO4 + Ba(OH)2 -> BaSO4 + Mg(OH)2

Mg(OH)2 -> MgO + H2O

MgO + 2HCl -> MgCl2 + H2

MgCl2 + NaOH -> Mg(OH)2 + NaCl

Answer 3

3.a ) \(2Na+\dfrac{1}{2}O_2-^{t^o}\rightarrow Na_2O\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)

\(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2+H_2O\)

b)\(Mg+\dfrac{1}{2}O_2-^{t^o}\rightarrow MgO\)

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

\(MgSO_4+2NaOH\rightarrow Mg\left(OH\right)_2+Na_2SO_4\)

\(Mg\left(OH\right)_2-^{t^o}\rightarrow MgO+H_2O\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

\(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\)

Answer 4

1.a/\(n_{HCl}=0,1.2=0,2\left(mol\right)\)

\(2HCl+Ba\left(OH\right)_2\rightarrow BaCl_2+H_2O\)

\(TheoPT:n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\)

\(\Rightarrow V=\dfrac{0,1}{0,2}=0,5M\)

b)\(TheoPT:n_{BaCl_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\)

\(\Rightarrow CM_{BaCl_2}=\dfrac{0,1}{0,1+0,2}=0,33M\)

Answer 5

bài 1

a)n_HCl=0,1.2=0,2(mol)

PTHH:2HCl+Ba(OH)₂-->BaCl₂+2H₂O(1)

           0,2      0,1            0,1                 (mol)

từ pt (1)-->n_Ba(OH)2=0,1(mol)

-->V_Ba(OH)2=0,1.0,2=0.02(lít)

b)từ pt(1)-->n_BaCl2=0,1(mol)

Vdd sau pư =0,1+0,02=0,12(lít)

--->C_M=0,1/0,12=5/6(M)