PT: \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

Ta có: \(n_{H_2SO_4}=0,25.1=0,25\left(mol\right)\)

a, Theo PT: \(n_{NaOH}=2n_{H_2SO_4}=0,5\left(mol\right)\Rightarrow V_{NaOH}=\dfrac{0,5}{2}=0,25\left(l\right)\)

b, Theo PT: \(n_{Na_2SO_4}=n_{H_2SO_4}=0,25\left(mol\right)\)

\(\Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,25}{0,25+0,25}=0,5\left(M\right)\)

\(C_{M_{H_2SO_4}}=\dfrac{n}{V}=\dfrac{0,1}{0,25}=0,4M\)

a, \(C_{M_{H_2SO_4}}=\dfrac{0,1}{0,25}=0,4\left(M\right)\)

b, \(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)

\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,5}{0,4}=1,25\left(M\right)\)

\(a)C_{M\left(H_2SO_4\right)}=\dfrac{0,1}{0,25}=0,4M\\ b)C_{M\left(NaOH\right)}=\dfrac{\dfrac{20}{40}}{0,4}=1,25M\)

\(a,n_{H_2SO_4}=0,1mol\)

\(V_{H_2SO_4}=0,25l\left(250ml=0,25l\right)\\ C_{M_{H_2SO_4}}=\dfrac{n}{V}=\dfrac{0,1}{0,25}=0,4M\)

`b,` \(n_{NaOH}=\dfrac{m}{M}=\dfrac{20}{40}=0,5mol\)

\(V_{NaOH}=0,4l\left(400ml=0,4l\right)\)

\(C_{M_{NaOH}}=\dfrac{n}{V}=\dfrac{0,5}{0,4}=1,25M\)

Bài 1 :

m_NaNO3 = C% * m_{dung dịch} / 100% = 5,1 * 200 / 100 =10,2 ( g )

n _{ct NaNO3} = m/M = 10,2 / 85 = 0,12 ( mol )

Câu 1:

Áp dụng công thức :

\(C\%=\dfrac{m_{ct}.100\%}{m_{dd}}\)

=> m_NaNO3=\(\dfrac{C\%.m_{dd}}{100\%}=\dfrac{5,1.200}{100}=10,2\left(g\right)\)

=> n_NaNO3=m/M=10,2/85=0,12(mol)

câu 2 :

V_{d d} =250ml=0,25(lít)

n_K2SO4=m/M=26,1/174=0,15(mol)

=> C_M=\(\dfrac{n}{V}=\dfrac{0,15}{0,25}=0,6\left(M\right)\)

khối lượng chất tan có trong 200đ dung dịch NaNO3​ là

mct =\(\dfrac{C\%.mdd}{100\%}\)

mct =\(\dfrac{5,1.200}{100}\)=10,2 (g)

số mol chất tan là

n=\(\dfrac{m}{M}\)

nct=\(\dfrac{10,2}{85}\)= 0,12 mol

1: NaOH+HCl->NaCl+H2O

0,375        0,375

\(V_{HCl}=0.375\cdot22.4=8.4\left(lít\right)\)

\(C_{M\left(NaCl\right)}=\dfrac{0.375}{8.4+0.25}=\dfrac{15}{346}\)

\(n_{NaOH}=1,5.0,25=0,375\left(mol\right)\)

Pt : \(NaOH+HCl\rightarrow NaCl+H_2O\)

a) Theo Pt : \(n_{NaOH}=n_{HCl}=n_{NaCl}=0,375\left(mol\right)\)

\(V_{ddHCl}-\dfrac{0,375}{1,5}=0,25\left(l\right)\)

b) \(C_{MNaCl}=\dfrac{0,375}{0,25}=1,5\left(M\right)\)

 Chúc bạn học tốt 

\(a.n_{NaOH}=1,5.0,25=0,375mol\\ NaOH+HCl\rightarrow NaCl+H_2O\\ n_{NaOH}=n_{HCl}=n_{NaCl}=0,375mol\\ V_{HCl}=\dfrac{0,375}{1,5}=0,25l\\ b.C_{M\left(NaCl\right)}=\dfrac{0,375}{0,25+0,25}=0,75M\)

Bài 1:

\(a.n_{NaOH\left(tổng\right)}=0,05.1+0,2.0,2=0,09\left(mol\right)\\ V_{ddNaOH\left(tổng\right)}=50+200=250\left(ml\right)=0,25\left(l\right)\\ C_{MddNaOH\left(cuối\right)}=\dfrac{0,09}{0,25}=0,36\left(M\right)\\ b.n_{HCl}=0,5.0,02=0,01\left(mol\right)\\ n_{H_2SO_4}=0,08.0,2=0,016\left(mol\right)\\ V_{ddsau}=20+80=100\left(ml\right)=0,1\left(l\right)\\ C_{MddH_2SO_4}=\dfrac{0,016}{0,1}=0,16\left(M\right)\\ C_{MddHCl}=\dfrac{0,01}{0,1}=0,1\left(M\right)\)

Bài 2:

\(a.m_{H_2SO_4}=29,4.10\%=2,94\left(g\right)\\ b.n_{H_2SO_4}=\dfrac{2,94}{98}=0,03\left(mol\right)\\ n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Vì:\dfrac{0,01}{1}< \dfrac{0,03}{1}\Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(dư\right)}=0,03-0,01=0,02\left(mol\right)\\ m_{H_2SO_4\left(dư\right)}=0,02.98=1,96\left(g\right)\\ n_{H_2}=n_{Fe}=0,01\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,01.22,4=0,224\left(l\right)\)

\(H_2SO_4+2NaOH->Na_2SO_4+2H_2O\\ H_2SO_4+Na_2CO_3->Na_2SO_4+CO_2+H_2O\\ n_{Na_2CO_3}=0,1mol=n_{H_2SO_4dư}\\ n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}.0,15.2=0,15mol\\ C_{M\left(H_2SO_4\right)}=\dfrac{0,15+0,1}{0,25}=1\left(M\right)\\ m_{Na_2SO_4}=142\left(0,15+0,1\right)=35,5g\)