NaOH +HCl--->NaCl +H2O
Ta có
n\(_{NaOH}=\frac{50}{40}=1,25\left(mol\right)\)
n\(_{HCl}=\frac{36,5}{36,5}=1\left(mol\right)\)
=> NaOH dư
Theo pthh
n\(_{NaCl}=n_{HCl}=1\left(mol\right)\)
m\(_{NaCl}=1.58,5=58,5\left(g\right)\)
Theo pthh
n\(_{NaOH}=n_{HCl}=0,1\left(mol\right)\)
n\(_{NaOH}dư=1,25=1=0,25\left(mol\right)\)
m\(_{NaOH}\)dư =0,25.40=10(g)
Ta có : n NaOH = 1,25
nHCl = 1 (mol)
PTHH : NaOH + HCl--->NaCl+H2O
=> n NaOH dư
=> n NaCl = 1 (mol)
=> m NaCl = 1.58,5=58,5
=> nNAOH = 0,1 (mol)
=> n NaOH dư = 0,25
=> m NaOH = 10