Đề kiểm tra cuối kì I: đề 2

\(M_{Al_2O_3}=27.2+16.3=102\left(DvC\right)\\ \%Al=\dfrac{27.2}{102}.100\%=52,94\%\)

hmm k có đáp án nào hết

27,54%

27,54%

Câu 1:

2Mg + O₂ \(\underrightarrow{t^o}\) 2MgO

\(n_{Mg}=\dfrac{7,2}{24}=0,3mol\)

\(\Rightarrow n_{MgO}=\dfrac{0,3.2}{2}=0,3mol\)

\(m_{MgO}=0,3.40=12g\)

Câu 2:

CaCO₃ \(\rightarrow\) CO₂\(\uparrow\) + CaO

\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1mol\)

\(\Rightarrow m_{CaCO_3}=0,1.100=10g\)

Câu 3:

Ba + 2HCl \(\rightarrow\) BaCl₂ + H₂

\(n_{BaCl_2}=\dfrac{4,16}{208}=0,02mol\)

\(\Rightarrow n_{HCl}=0,02.2=0,04mol\)

\(m_{HCl}=0,04.36,5=1,46g\)

Câu 4:

2Al + 3H₂SO₄ \(\rightarrow\) Al₂(SO₄)₃ + 3H₂

\(n_{H_2SO_4}=\dfrac{19,6}{98}=0,2mol\)

\(n_{H_2}=\dfrac{0,2.3}{3}=0,2mol\\ V_{H_2}=0,2.22,4=4,48l\)

chỉ cân câu 1 câu3 4 thôi nha

nếu đc thêm chú thích giúp tui một ti:)))

1)

\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\\ n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\\ n_{MgO}=n_{Mg}=0,3\left(mol\right)\\ m_{MgO}=0,3.40=12\left(g\right)\)

2)

\(CaCO_3\rightarrow\left(t^o\right)CaO+CO_2\uparrow\\ n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ m_{CaCO_3}=0,1.100=10\left(g\right)\)

3)

\(Ba+2HCl\rightarrow BaCl_2+H_2\uparrow\\ n_{BaCl_2}=\dfrac{4,16}{208}=0,02\left(mol\right)\\ n_{HCl}=0,02.2=0,04\left(mol\right)\\ m_{HCl}=0,04.36,5=1,46\left(g\right)\)

4)

\(n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\\ n_{H_2}=n_{H_2SO_4}=0,2\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)

\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ CaCO_3\rightarrow\left(t^o\right)CaO+CO_2\uparrow\\ n_{CaCO_3}=n_{CO_2}=0,1\left(mol\right)\\ m_{CaCO_3}=0,1.100=10\left(g\right)\)

\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ CaCO_3\rightarrow\left(t^o\right)CaO+CO_2\uparrow\\ n_{CaCO_3}=n_{CO_2}=0,1\left(mol\right)\\ m_{CaCO_3}=0,1.100=10\left(g\right)\)

\(PTHH:CaCO_3\underrightarrow{t^o}CO_2+CaO\)(pthh là CaO ko phải H₂O nhé)

\(n_{CO_2}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(Theo.PTHH:n_{CaCO_3}=n_{CO_2}=0,1\left(mol\right)\\ m_{CaCO_3}=n.M=0,1.100=10\left(g\right)\)

\(M+nHCl\rightarrow MCl_n+nH_2\)

Mg + 2HCl → MgCl2 + H2

hentaiz .com

\(n_{CO_2}=\dfrac{m}{M}=\dfrac{2,2}{44}=0,05\left(mol\right)\\ V_{O_2}=n.22,4=0,05.22,4=1,12\left(l\right)\\ n_{O_2}=\dfrac{m}{M}=\dfrac{4,8}{32}=0,15\left(mol\right)\\ V_{O_2}=n.22,4=0,15.22,4=3,36\left(l\right)\)

\(V_{CO_2}=\left(\dfrac{2,2}{44}\right).22,4=1,12l\)

\(V_{O_2}=\left(\dfrac{4,8}{32}\right).22,4=3,36l\)

\(n_{Co_2}=\dfrac{3,36}{22,4}=0,15mol\)

\(n_{N_2}=\dfrac{44,8}{22,4}=2mol\)

M_X = 0,552.29 = 16 (g/mol)

\(m_C=\dfrac{16.75}{100}=12\left(g\right)=>n_C=\dfrac{12}{12}=1\left(mol\right)\)

\(m_H=16-12=4\left(g\right)=>n_H=\dfrac{4}{1}=4\left(mol\right)\)

=> CTHH: CH₄

Phương trình hóa học:

C + O₂ → CO₂

Theo ĐLBTKL, ta có:

mC+mO2=mCO2⇒12+mO2=44⇒mO2=32mC+mO2=mCO2⇒12+mO2=44⇒mO2=32

Vậy, khối lượng O₂ đã tham gia phản ứng là 32kg.

44kg

Áp dụng định luân bảo toàn khối lượng 

m C+mO2=m CO2

=>m CO2=12+32=44kg

sao mik đưa hình ảnh lên mà k dc?

\(m_C=12\cdot\left(100-1.5-0.5\right)\%=11.76\left(kg\right)\)

\(n_C=\dfrac{11.76}{12}=0.98\left(kmol\right)\)

\(m_S=12\cdot0.5=6\left(kg\right)\)

\(n_S=\dfrac{6}{32}=0.1875\left(kmol\right)\)

\(S+O_2\underrightarrow{t^0}SO_2\)

\(C+O_2\underrightarrow{t^0}CO_2\)

\(V_{O_2}=\left(0.1875+0.98\right)\cdot22.4=26.152\left(kl\right)=26125\left(l\right)\)