Ta có: \(C_{\%_{NaOH}}=\dfrac{m_{NaOH}}{400}.100\%=4\%\)

=> \(m_{NaOH}=16\left(g\right)\)

=> \(n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\)

PTHH: HCl + NaOH ---> NaCl + H₂O

Theo PT: \(n_{NaCl}=n_{NaOH}=0,4\left(mol\right)\)

=> \(m_{NaCl}=0,4.58,5=23,4\left(g\right)\)

a,Gọi a là số mol NaOH b là số mol KOH

\(PTHH:NaOH+HCl\rightarrow NaCl+H2O\)

\(KOH+HCl\rightarrow KCl+H2O\)

Giải HPT:

\(\left\{{}\begin{matrix}\text{40x + 56y = 3,04g}\\\text{58,5x + 74,5y = 4,15 g}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\text{x= 0.02 mol }\\\text{y=0.04 mol}\end{matrix}\right.\)

\(\Rightarrow\text{mNaOH=0.02.40=0.8 g}\)

\(\Rightarrow\text{mKOH= 0.04.56 = 2,24 g}\)

b,\(\text{2NAOH + H2SO4 => NA2SO4 +2 H2O}\)

\(\text{mH2SO4=0.01.98=0.98 g}\)

\(\text{2KOH +H2SO4 =>K2SO4 + 2H2O}\)

\(\text{mH2SO4=0.02.98=1.96 g}\)

\(\text{mH2SO4=1.96+0.98=2.94 g}\)

\(\text{mddH2SO4=2,94.100:20=14,7 g}\)

a: \(HCl+NaOH\rightarrow NaCl+H_2O\)

\(n_{HCl}=0.3\cdot1=0.3\left(mol\right)\)

\(n_{NaOH}=0.2\cdot0.5=0.1\left(mol\right)\)

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

Lập tỉ lệ : \(\dfrac{0.3}{1}>\dfrac{0.1}{1}\Rightarrow HCldư\)

\(m_{NaCl}=0.1\cdot58.5=5.85\left(g\right)\)

HCl+NaOH->NaCl+H2O

         0,1---0,1

n NaOH=0,5.0,2=0,1 mol

=>m NaCl=0,1.58,5=5,85g

HCl+NaOH->NaCl+H2O

         0,1---0,1

n NaOH=0,5.0,2=0,1 mol

=>m NaCl=0,1.58,5=5,85g

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

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

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

NaOH có nồng độ bn vậy bạn

\(a,HCl+NaOH\rightarrow NaCl+H_2O\\ b,n_{HCl}=\dfrac{3,65}{36,5}=0,1mol\\ n_{NaOH}=0,2.1=0,2mol\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,2}{1}\Rightarrow NaOH.dư\\ n_{NaCl}=n_{NaOH,pư}=n_{HCl}=0,1mol\\ m_{NaOH,dư}=\left(0,2-0,1\right).40=4g\\ c,m_{NaCl}=0,1.58,5=5,85g\)