Đáp án:
a, Fe+H2SO4→FeSO4+H2
b, $n_{Fe}=\frac{11,2}{56}=0,2$ mol
=> $n_{H_{2}}=n_{H_{2}SO_{4}}=n_{Fe}=0,2$ mol
=> $V_{H_{2}(đktc)}=22,4.0,2=4,48l$
c, $CM_{H_{2}SO_{4}}=\frac{n_{H_{2}SO_{4}}}{V_{ddH_{2}SO_{4}}}= 0,2/0,2=1M$
d, $n_{H_{2}SO_{4}}= 0,1.1 = 0,1$ mol
PTHH:
Ca(OH)2+H2SO4→2H2O+CaSO4(
=> $n_{Ca(OH)_{2}}=n_{H_{2}SO_{4}}=0,1$ mol
=> $V_{ddCa(OH)_{2}}=0,1/2=0,05l=50ml$