a)\(n_{NaOH}:\dfrac{10}{40}=0,25\left(mol\right)\)
\(n_{HNO_3}:\dfrac{20}{63}\left(mol\right)\)
\(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)
1....................1..................1......................(mol)
0,25...............0,25..............0,25.................(mol)
->\(HNO_3\)dư
=> Dung dịch sau phản ứng có axit
b)\(m_{NaNO_3}:85.0,25=21,25\left(g\right)\)
\(m_{HNO_3}dư\):\(63.\left(\dfrac{20}{63}-0,25\right)=4,25\left(g\right)\)