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

$\text{Câu B:}$

$\text{mH2=0,5×2= 1 g}$

$\text{mCO2=0,75×44= 33 g}$

$\text{mN2=0,25×28= 7 g}$

$\text{⇒mHH=1+33+7= 41 g}$

$\text{VH2=0,5×22,4= 11,2 lít}$

$\text{VCO2=0,75×22,4= 16,8 lít}$

$\text{VN2=0,25×22,4= 5,6 lít}$

$\text{⇒VHH=11,2+16,8+5,6= 33,6 lít}$

$\text{Câu C:}$

$\text{MA=$\dfrac{28}{0,5}$= 56 g/mol}$

$\text{Vậy A là sắt, kí hiệu là Fe}$

$\text{Câu D:}$

$\text{mAl2(SO4)3=0,2×342= 68,4 g}$

$\text{Số PT=0,2×6.10²³= 1,2.10²³}$

$\text{Câu E:}$

$\text{MFe2O3=56×2+16×3= 160 g/mol}$

$\text{%mFe=$\dfrac{56×2}{160}$×100= 70%}$

$\text{%mO=$\dfrac{16×3}{160}$×100= 30%}$

$\text{MMgO=24+16= 40g/mol}$

$\text{%mMg=$\dfrac{24}{40}$×100= 60%}$

$\text{%mO=$\dfrac{16}{40}$×100= 40%}$

$\text{MCa(NO3)2=40+(14+16×3)×2= 164 g/mol}$

$\text{%mCa=$\dfrac{40}{164}$×100= 24,39%}$

$\text{%mN=$\dfrac{14×2}{164}$×100= 17,07%}$

$\text{%mCa=$\dfrac{16×3×2}{164}$×100= 58,54%}$

$\text{Câu F:}$

$\text{MCO=16+12= 28 g/mol}$

$\text{%mC=$\dfrac{12}{28}$×100= 42,86%}$

$\text{MCO2=12+16×2= 44 g/mol}$

$\text{%mC=$\dfrac{12}{44}$×100= 27,27%}$

$\text{MCH4=12+1×4= 16 g/mol}$

$\text{%mC=$\dfrac{12}{16}$×100= 75%}$

$\text{Ta có:}$

$\text{27,27%<42,86%<75%}$

$\text{⇒Hợp chất CH4 có tỉ lệ cacbon cao nhất}$

$\text{Câu G:}$

$\text{Gọi CTDC của hợp chất: CxHy}$

$\text{Ta có tỉ lệ: x:y=$\dfrac{75}{12}$:$\dfrac{25}{2}$=6,25:12,5=1:2}$

$\text{Chọn x=1, y=2}$

$\text{Vậy CTHH của hợp chất: CH2}$

$\text{Câu i:}$

$\text{nN2=$\dfrac{9.10²³}{6.10²³}$= 1,5 mol}$

$\text{⇒mN2=1,5×28= 42 g}$3

$\text{Câu j:}$

$\text{nFe=$\dfrac{1,8.10²³}{6.10²³}$= 0,3 mol}$

$\text{nH2O=$\dfrac{24.10²³}{6.10²³}$= 4 mol}$

$\text{Câu t:}$

$\text{nCl2=$\dfrac{142}{71}$= 2 mol}$

$\text{⇒VCl2=2×22,4= 44,8 lít}$

$\text{nCO2=$\dfrac{3,01.10²³}{6.10²³}$= 0,5 mol}$

$\text{⇒VCO2=0,5×22,4= 11,2 lít}$

$\text{Chúc bạn học tốt}$