Câu 10: $D$
$n_{KOH}=n_{NaOH}=0,5.0,1=0,05(mol)$
$\to m_{KOH}=0,05.56=2,8g=2800mg$
Suy ra $1g$ CB cần $\dfrac{2800}{100}=28mg$ $KOH$ trung hoà
Câu 11: $B$
$4g$ CB cần $7.4=28mg$ $KOH$
$n_{NaOH}=n_{KOH}=\dfrac{28}{56}=0,5(mmol)=5.10^{-4}(mol)$
$\to m=5.10^{-4}.40=0,02g$
Câu 12: $A$
$28g$ CB cần $28.6=168mg$ $KOH$
$n_{KOH}=\dfrac{168}{56}=3(mmol)=0,003(mol)$
$\to n_{Ba(OH)_2}=\dfrac{n_{KOH}}{2}=0,0015(mol)$
$\to m=0,0015.171=0,2565g$
Câu 13: $D$
$15g$ CB cần $15.7=105mg=0,105g$ $KOH$
$n_{NaOH}=n_{KOH}=\dfrac{0,105}{56}=1,875.10^{-3}(mol)$
$\to m=1,875.10^{-3}.40=0,075g$
