\(n_{H_3PO_4}=0.2\cdot0.5=0.1\left(mol\right)\)
\(3NaOH+H_3PO_4\rightarrow Na_3PO_4+3H_2O\)
\(0.3..............0.1\)
\(m_{dd_{NaOH}}=\dfrac{0.3\cdot40}{40\%}=30\left(g\right)\)
\(V_{dd_{NaOH}}=\dfrac{30}{1.2}=25\left(ml\right)\)
PTHH: \(3NaOH+H_3PO_4\rightarrow Na_3PO_4+3H_2O\)
Ta có: \(n_{H_3PO_4}=0,5\cdot0,2=0,1\left(mol\right)\) \(\Rightarrow n_{NaOH}=0,3\left(mol\right)\)
\(\Rightarrow m_{ddNaOH}=\dfrac{0,3\cdot40}{40\%}=30\left(g\right)\) \(\Rightarrow V_{ddNaOH}=\dfrac{30}{1,2}=25\left(ml\right)\)
3NaOH+H3PO4->Na3PO4+3H2O
Số mol h3po4=0,2.0,5=0,1mol
=> n(NaOH)=0,1.3=0,3mol
mNaOH=0,3.40=12g
mddNaOH=12.100/40=30g
V=m/d=30/1,2=25ml
nH3PO4= 0.5 x 0.2 = 0.1 (mol)
3NaOH + H3PO4 --------> Na3PO4 + 3H2O
(mol) 0.3 <----- 0.1
=> mNaOH = 0.3 x 40 = 12 (g)
=> mddNaOH = (12/40)x100 = 30 (g)
=> VddNaOH = \(\dfrac{30}{1.2}\)= 25 (ml)