1.

Trong 1 mol $C_{12}H_{22}O_{11}$ có 12 nguyên tử C, 22 nguyên tử H và 11 nguyên tử O

2.

$a,PTK_A=PTK_{H_2}.40=2.40=80(đvC)\\ b,CTTQ_A:XO_3\\ \Rightarrow NTK_X+3NTK_O=80\\ \Rightarrow NTK_X+3.16=80\\ \Rightarrow NTK_X=32(đvC)\\ \Rightarrow \text{X là lưu huỳnh (S)}$

3.

$a,PTK_{CO_2}=12+16.2=44(đvC)\\ b,PTK_{HNO_3}=1+14+16.3=63(đvC)\\ c,PTK_{KClO_3}=39+35,5+16.3=122,5(đvC)$

4.

$a,PTK_{hợp chất}=31.PTK_{H_2}=31.2=62(đvC)\\ b,CTTQ:X_2O\\ \Rightarrow PTK_{X_2O}=2NTK_X+NTK_O=62\\ \Rightarrow 2NTK_X+16=62\\ \Rightarrow NTK_X=23(đvC)\\ \text{X là natri (Na)}$

5.

a, magie + khí oxi  ---$t^0$--> magie oxit 

b, kẽm + axit clohidric ------> kẽm clorua + khí hiđro

c, bari clorua + axit sunfuric ------> bari sunfat + axit clohidric

6.

$a,2Fe+3Cl_2\xrightarrow{t^o}2FeCl_3\\ b,2Al(OH)_3\xrightarrow{t^o}Al_2O_3+3H_2O\\ c,3Ca(OH)_2+2H_3PO_4\to Ca_3(PO_4)_2+6H_2O\\ d,2ZnS+3O_2\xrightarrow{t^o}2ZnO+2SO_2$

7.

$a,4Na+O_2\xrightarrow{t^o}2Na_2O\\ b,2Na_3PO_4+3BaCl_2\to Ba_3(PO_4)_2+6NaCl\\ c,Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O\\ d,Fe_2O_3+3CO\xrightarrow{t^o}2Fe+3CO_2$

$\text{Đáp án+Giải thích các bước giải:}$

$\text{II Tự luận:}$

$\text{Câu 1:}$

$\text{Trong hợp chất $C_{12}H_{22}O_{11}$ có: 12 mol nguyên tử C, 22 mol nguyên tử H, 11 mol nguyên tử O}$

$\text{$n_{C}$=12×1= 12 mol}$

$\text{$n_{H}$=22×1= 22 mol}$

$\text{$n_{O}$=11×1= 11 mol}$

$\text{Câu 2:}$

$\text{Gọi CTDC: $XO_{3}$}$

$\text{A.}$

$\text{PTK=2×40= 80 đvC}$

$\text{B.}$

$\text{X=80-$O_{3}$=80-48= 32 đvC}$

$\text{Vậy X là lưu huỳnh, kí hiệu là S}$

$\text{Câu 3:}$

$\text{a) PTK=12+16×2= 44 đvC}$

$\text{b) PTK=1+14+16×3= 63 đvC}$

$\text{c) PTK=39+35,5+16×3= 122,5 đvC}$

$\text{Câu 4:}$

$\text{Gọi CTDC: $X_{2}O$}$

$\text{A.}$

$\text{PTK=2×31= 62 đvC}$

$\text{B.}$

$\text{$X_{2}$=62-O=62-16= 46 }$

$\text{⇒ X=$\dfrac{46}{2}$= 23 đvC}$

$\text{Vậy X là Natri, kí hiệu là Na}$

$\text{Câu 5:}$

$\text{a) 2Mg+$O_{2}$$\xrightarrow{t^o}$2MgO}$

$\text{b) Zn+2HCl→$ZnCl_{2}$+$H_{2}$}$

$\text{c) $BaCl_{2}$+$H_{2}SO_{4}$→$BaSO_{4}$+2HCl}$

$\text{Câu 6:}$

$\text{a) 2Fe+$3Cl_{2}$→$2FeCl_{3}$}$

$\text{b) $2Al(OH)_{3}$→$Al_{2}O_{3}$+$3H_{2}O$}$

$\text{c) $3Ca(OH)_{2}$+$2H_{3}PO_{4}$→$Ca_{3}(PO_{4})_{2}$+$6H_{2}O$}$

$\text{d) 2ZnS+$3O_{2}$$\xrightarrow{t^o}$2ZnO+$2SO_{2}$}$

$\text{Câu 7:}$

$\text{a) 4Na+$O_{2}$$\xrightarrow{t^o}$$2Na_{2}O$}$

$\text{b) $2Na_{3}PO_{4}$+$3BaCl_{2}$→6NaCl+$Ba_{3}(PO_{4})_{2}$}$

$\text{c) $Al_{2}O_{3}$+$3H_{2}SO_{4}$→$Al_{2}(SO_{4})_{3}$+$3H_{2}O$}$

$\text{d) $Fe_{2}O_{3}$+3CO$\xrightarrow{t^o}$2Fe+$3CO_{2}$}$

$\text{Chúc bạn học giỏi hóa}$