Ca+2H2O⇒Ca(OH)2+H2↑        (1)

CaO+H2O⇒Ca(OH)2                (2)

nCa=4÷40=0.1(mol)

nCaO=5.6÷(40+16)=0.1(mol)

+)nH2=nCa=0.1(mol)  ⇒V(H2)=0.1×22.4=2.24(l)

+) nCa(OH)2(1)=nCa=0.1(mol)        ⇒mCa(OH)2(1)=0.1×[40+(16+1)×2]=7.4(g)

    nCa(OH)2(2)=nCaO=0.1(mol)     ⇒mCa(OH)2(2)=0.1×[40+(16+1)×2]=7.4(g)

⇒∑mCa(OH)2=mCa(OH)2(1)+mCa(OH)2(2)=7.4+7.4=14.8(g)

Đáp án:

\({{\text{V}}_{{H_2}}}{\text{ = 2}}{\text{,24 lít; }}{{\text{m}}_{Ca{{(OH)}_2}}} = 14,8{\text{ gam}}\)

Giải thích các bước giải:

 Ta có:

\(n{  _{Ca}} = \frac{4}{{40}} = 0,1{\text{mol; }}{{\text{n}}_{CaO}} = \frac{{5,6}}{{40 + 16}} = 0,1{\text{ mol}}\)

Phản ứng xảy ra:

\(Ca + 2{H_2}O\xrightarrow{{}}Ca{(OH)_2} + {H_2}\)

\(CaO + {H_2}O\xrightarrow{{}}Ca{(OH)_2}\)

\( \to {n_{{H_2}}} = {n_{Ca}} = 0,1{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)

\({n_{Ca{{(OH)}_2}}} = {n_{Ca}} + {n_{CaO}} = 0,1 + 0,1 = 0,2{\text{ mol}} \to {{\text{m}}_{Ca{{(OH)}_2}}} = 0,2.(40 + 17.2) = 14,8{\text{ gam}}\)