Ta có: $100ml=0,1(l)$
a,$nCaCO_3=10/100=0,1(mol)$
$nH_2SO_4=0,1.1,5=0,15(mol)$
$mddH_2SO_4=1,1.100=110(g)$
$CaCO_3+H_2SO_4→CaSO_4+H_2O+CO_2↑$
Vì $0,1/1<0,15/1$ nên $H_2SO_4$ dư 0,05 mol hay dư 0,05.98=4,9 g
$nCaSO_4=0,1(mol)$
$⇒mCaSO_4=0,1.136=13,6(g)$
$∑mdd=110+10-0,1.44=115,6(g)$
⇒%$CaSO_4=13,6.100/115,6≈11,76$%
%$H_2SO_4 dư=4,9.100/115,6≈4,24$%
b,$H_2SO_4+2NaOH→Na_2SO_4+2H_2O$
$⇒nNaOH cần=0,1(mol)$
$⇒V_{ddNaOH4M}=0,1/4=0,025(l)=25(ml)$
$⇒m_{ddNaOH4M}=25.1,2=30(g)$