\(m_{NaOH}=\frac{50.10}{100}=5\left(g\right)\Rightarrow n_{NaOH}=\frac{5}{40}=0,125\left(mol\right)\)
\(PTHH:\text{ }H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
\(Theo\text{ PT: }n_{H_2SO_4}=\frac{1}{2}n_{NaOH}=\frac{1}{2}.0,125=0,0625\left(mol\right)\)
\(Vdd_{H_2SO_4}=\frac{n_{H_2SO_4}}{C_M}=\frac{0,0625}{0,5}=0,125\left(l\right)=125\text{ }ml\)
\(m_{NaOH}=\frac{50.10}{100}\)=5 g
\(n_{NaOH}=\frac{5}{40}=0,125\left(mol\right)\)
PTHH: H2SO4+2NaOH\(\rightarrow\)Na2SO4+2H2O
0,0625 \(\leftarrow\)0,125( mol)
VẬY \(V_{H2SO4}=\frac{0,0625}{0,5}=0,125\left(l\right)\)
ta có: 1l=1000ml\(\Rightarrow\) 0,125(l)=125ml
chúc bạn học tốt like nha
nNaOH=\(\frac{50\cdot10}{100\cdot40}=0,125mol\)
PTHH: H2SO4 + 2NaOH----> Na2SO4 + 2H2o
Gọi V ddH2So4 là x(\(l\))----> nH2SO4=0,5x\(mol\)
Theo PT : nH2SO4=1/2nNaOh-----> 0,5x=\(\frac{1}{2}\).0.125=0,0625
----> x=\(\frac{0,0625}{0,5}=0,125l\)l= 125 ml
Vậy VddH2SO4= 125ml