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![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(n_{NaCl}=n_{NaOH}=0,4\left(mol\right)\)
=> \(m_{NaCl}=0,4.58,5=23,4\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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,\(\text{2NAOH + H2SO4 => NA2SO4 +2 H2O}\)
\(\text{mH2SO4=0.01.98=0.98 g}\)
\(\text{2KOH +H2SO4 =>K2SO4 + 2H2O}\)
\(\text{mH2SO4=0.02.98=1.96 g}\)
\(\text{mH2SO4=1.96+0.98=2.94 g}\)
\(\text{mddH2SO4=2,94.100:20=14,7 g}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{CaCO3}=\dfrac{30}{100}=0,3\left(mol\right)\)
\(n_{NaOH}=2.0,25=0,5\left(mol\right)\)
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
mNaOH = \(\dfrac{C\%\times m_{ddNaOH}}{100\%}\)=\(\dfrac{4\%\times400}{100\%}\) (g)
nNaOH = \(\dfrac{16}{40}\)= 0,4 (mol)
mNaCl = 0,4\(\times\)58,5 = 23,4 (g)