m_{dd HNO3}=15.1,4=21(g)

m_HNO3= \(\frac{ }{\frac{\frac{ }{ }\frac{ }{ }\frac{ }{ }21.60\%}{:100\%}}\)21.60%/100%=12,6(g)

n_HNO3=12,6/63=0,2(mol)

PTHH: NaOH + HNO₃ → NaNO₃ + H₂O

0,2mol<--- 0,2mol

100ml=0,1l

C_M=0,2/0,1=2M

2.

PTHH: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

0,2mol → 0,1mol

m_H2SO4= 0,1.98=9,8(g)

m_{dd H2SO4}= 9,8*100%/49%=20(g)