a, Có: \(n_{O_2}=\dfrac{21,28}{22,4}=0,95\left(mol\right)\)

Theo ĐLBT KL, có: m + m_O2 = m_CO2 + m_H2O

⇒ m = 28,6 + 14,4 - 0,95.32 = 12,6 (g)

b, Có: \(n_{CO_2}=\dfrac{28,6}{44}=0,65\left(mol\right)\)

\(n_{H_2O}=\dfrac{14,4}{18}=0,8\left(mol\right)\)

BTNT O, có: n_CO + 2n_O2 = 2n_CO2 + n_H2O

⇒ n_CO = 0,65.2 + 0,8 - 0,95.2 = 0,2 (mol)

⇒ m_CO = 0,2.28 = 5,6 (g)

\(\Rightarrow\%m_{CO}=\dfrac{5,6}{12,6}.100\%\approx44,44\%\)

Bạn tham khảo nhé!

\(a.n_{CO_2}=0,5\left(mol\right)\Rightarrow n_C=0,5\left(mol\right)\\ n_{H_2O}=0,5\left(mol\right)\Rightarrow n_H=1\left(mol\right)\\ \Rightarrow n_O=\dfrac{15-0,5.12-1.1}{16}=0,5\left(mol\right)\\ ĐặtCTPTcủaAlà:C_xH_yO_z\left(x,y,z>0\right)\\ Tacó:x:y:z=0,5:1:0,5=1:2:1\\ \Rightarrow CTđơngiảnnhất:\left(CH_2O\right)_n\\ MàM_A=60\\ \Rightarrow\left(12+2+16\right).n=60\\ \Rightarrow n=2\\ VậyCTPTcủaA:C_2H_4O_2\\ b.VìAtácdụngvớiNa,CuO,NaOH\\ \Rightarrow AlàAxit\\ \Rightarrow CTCTcủaA:CH_3COOH\\ PTHH:CH_3COOH+Na\rightarrow CH_3COONa+\dfrac{1}{2}H_2\\2CH_3COOH +CuO\rightarrow\left(CH_3COO\right)_2Cu+H_2O\\ CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

Chúc em thi tốt !

câu b) lmj có Na(OH)₂ nhỉ

\(Tacó:m_{Fe_2O_3}:m_{CuO}=3:2\\ m_{Fe_2O_3}+m_{CuO}=40\\ \Rightarrow m_{Fe_2O_3}=24\left(g\right)\Rightarrow n_{Fe_2O_3}=0,15\left(mol\right)\\ \Rightarrow m_{CuO}=16\left(g\right)\Rightarrow n_{CuO}=0,2\left(mol\right)\\ Fe_2O_3+3H_2-^{t^o}\rightarrow2Fe+3H_2O\left(1\right)\\ CuO+H_2O-^{t^o}\rightarrow Cu+H_2O\left(2\right)\\TheoPT\left(1\right): n_{Fe}=2n_{Fe_2O_3}=0,3\left(mol\right)\\ \Rightarrow m_{Fe}=16,8\left(g\right)\\TheoPT\left(2\right): n_{Cu}=n_{CuO}=0,2\left(mol\right)\\ \Rightarrow m_{Cu}=0,2.64=12,8\left(g\right)\)

mFe =11,2g

Bài 1:

PTHH: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,3}{1}\) \(\Rightarrow\) Oxi còn dư, Hidro p/ứ hết

\(\Rightarrow n_{H_2O}=0,2\left(mol\right)\) \(\Rightarrow m_{H_2O}=0,2\cdot18=3,6\left(g\right)\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right);n_{CO}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ Fe_2O_3+3CO\rightarrow\left(t^o\right)2Fe+3CO_2\\ V\text{ì}:\dfrac{0,1}{1}>\dfrac{0,15}{3}\Rightarrow Fe_2O_3d\text{ư}\\ n_{Fe_2O_3\left(d\text{ư}\right)}=0,1-\dfrac{0,15}{3}=0,05\left(mol\right)\\ n_{Fe}=\dfrac{2}{3}.0,15=0,1\left(mol\right)\\ m=m_{r\text{ắn}}=m_{Fe_2O_3\left(d\text{ư}\right)}+m_{Fe}=0,05.160+0,1.56=13,6\left(g\right)\)

Khối lượng fe = 11,2 g