Câu hỏi của Mai Xuân

Question

Giúp em câu 18 vs ạ em cảm ơn

๖ۣۜDũ๖ۣۜN๖ۣۜG · Accepted Answer

$n_{H_2}=\dfrac{8,064}{22,4}=0,36\left(mol\right)$PTHH: 2R + 2nHCl --> 2RCln + nH2 $\dfrac{0,72}{n}$<-----------------0,36$m_{tăng}=m_R-m_{H_2}=\dfrac{0,72}{n}.M_R-0,36.2=19,44\left(g\right)$=> $M_R=28n\left(g/mol\right)$Xét n = 2 thỏa mãn => MR = 56 (g/mol)=> R là Fe$n_{Fe}=0,36\left(mol\right)$$n_{CO\left(bđ\right)}=\dfrac{16,128}{22,4}=0,72\left(mol\right)$$n_{Ba\left(OH\right)_2}=0,21.2=0,42\left(mol\right)$; $n_{BaCO_3}=\dfrac{59,1}{197}=0,3\left(mol\right)$TH1: Kết tủa không bị hòa tanPTHH: Ba(OH)2 + CO2 --> BaCO3 + H2O 0,3<----0,3 FexOy + yCO --to--> xFe + yCO2 $\dfrac{0,3x}{y}$<-0,3=> $\dfrac{0,3x}{y}=0,36\Rightarrow\dfrac{x}{y}=\dfrac{6}{5}\left(L\right)$TH2: Kết tủa bị hòa tan 1 phầnPTHH: Ba(OH)2 + CO2 --> BaCO3 + H2O 0,42---->0,42---->0,42 BaCO3 + CO2 + H2O --> Ba(HCO3)2 0,12-->0,12=> $n_{CO_2}=0,42+0,12=0,54\left(mol\right)$PTHH: FexOy + yCO --to--> xFe + yCO2 $\dfrac{0,54x}{y}$<-0,54=> $\dfrac{0,54x}{y}=0,36\Rightarrow\dfrac{x}{y}=\dfrac{2}{3}$=> CTHH: Fe2O3$n_{Fe_2O_3}=\dfrac{1}{2}.n_{Fe}=0,18\left(mol\right)\Rightarrow m=0,18.160=28,8\left(g\right)$Câu trả lời: $n_{H_2}=\dfrac{8,064}{22,4}=0,36\left(mol\right)$PTHH: 2R + 2nHCl --> 2RCln + nH2 $\dfrac{0,72}{n}$<-----------------0,36$m_{tăng}=m_R-m_{H_2}=\dfrac{0,72}{n}.M_R-0,36.2=19,44\left(g\right)$=> $M_R=28n\left(g/mol\right)$Xét n = 2 thỏa mãn => MR = 56 (g/mol)=> R là Fe$n_{Fe}=0,36\left(mol\right)$$n_{CO\left(bđ\right)}=\dfrac{16,128}{22,4}=0,72\left(mol\right)$$n_{Ba\left(OH\right)_2}=0,21.2=0,42\left(mol\right)$; $n_{BaCO_3}=\dfrac{59,1}{197}=0,3\left(mol\right)$TH1: Kết tủa không bị hòa tanPTHH: Ba(OH)2 + CO2 --> BaCO3 + H2O 0,3<----0,3 FexOy + yCO --to--> xFe + yCO2 $\dfrac{0,3x}{y}$<-0,3=> $\dfrac{0,3x}{y}=0,36\Rightarrow\dfrac{x}{y}=\dfrac{6}{5}\left(L\right)$TH2: Kết tủa bị hòa tan 1 phầnPTHH: Ba(OH)2 + CO2 --> BaCO3 + H2O 0,42---->0,42---->0,42 BaCO3 + CO2 + H2O --> Ba(HCO3)2 0,12-->0,12=> $n_{CO_2}=0,42+0,12=0,54\left(mol\right)$PTHH: FexOy + yCO --to--> xFe + yCO2 $\dfrac{0,54x}{y}$<-0,54=> $\dfrac{0,54x}{y}=0,36\Rightarrow\dfrac{x}{y}=\dfrac{2}{3}$=> CTHH: Fe2O3$n_{Fe_2O_3}=\dfrac{1}{2}.n_{Fe}=0,18\left(mol\right)\Rightarrow m=0,18.160=28,8\left(g\right)$

Kudo Shinichi · Answer

CTHH của oxit: `R_xO_y`$n_{CO\left(ban,đầu\right)}=\dfrac{16,128}{22,4}=0,72\left(mol\right)$$n_{H_2}=\dfrac{8,064}{22,4}=0,36\left(mol\right)$Ta có: $m_{tăng}=m_{KL}-m_{H_2}=19,44\left(g\right)$$\rightarrow m_{KL}=19,44+0,36.2=20,16\left(g\right)$PTHH: $2R+2nHCl\rightarrow2RCl_n+nH_2$ $\dfrac{0,72}{n}$<----------------------0,36$\rightarrow M_R=\dfrac{20,16}{\dfrac{0,72}{n}}=28n\left(g\text{/}mol\right)$Với `n = 2 -> M_R = 56` $(g/mol)$=> R là Fe$n_{Fe}=\dfrac{20,16}{56}=0,36\left(mol\right)$$n_{Ba\left(OH\right)_2}=0,21.2=0,42\left(mol\right)$$n_{BaCO_3}=\dfrac{59,1}{197}=0,3\left(mol\right)\ \xrightarrow[]{\text{BTNT Ba}}n_{Ba\left(HCO_3\right)_2}=0,42-0,3=0,12\left(mol\right)\ \xrightarrow[]{\text{BTNT C}}n_{CO_2}=0,12.2+0,3=0,54\left(mol\right)$So sánh: `0,54 < 0,72 => CO` dưPT rút gọn: $CO+O\rightarrow CO_2$ 0,54<-0,54CTHH: `Fe_xO_y``-> x : y = 0,36 : 0,54 = 2 : 3`CTHH: `Fe_2O_3`Câu trả lời: CTHH của oxit: `R_xO_y`$n_{CO\left(ban,đầu\right)}=\dfrac{16,128}{22,4}=0,72\left(mol\right)$$n_{H_2}=\dfrac{8,064}{22,4}=0,36\left(mol\right)$Ta có: $m_{tăng}=m_{KL}-m_{H_2}=19,44\left(g\right)$$\rightarrow m_{KL}=19,44+0,36.2=20,16\left(g\right)$PTHH: $2R+2nHCl\rightarrow2RCl_n+nH_2$ $\dfrac{0,72}{n}$<----------------------0,36$\rightarrow M_R=\dfrac{20,16}{\dfrac{0,72}{n}}=28n\left(g\text{/}mol\right)$Với `n = 2 -> M_R = 56` $(g/mol)$=> R là Fe$n_{Fe}=\dfrac{20,16}{56}=0,36\left(mol\right)$$n_{Ba\left(OH\right)_2}=0,21.2=0,42\left(mol\right)$$n_{BaCO_3}=\dfrac{59,1}{197}=0,3\left(mol\right)\ \xrightarrow[]{\text{BTNT Ba}}n_{Ba\left(HCO_3\right)_2}=0,42-0,3=0,12\left(mol\right)\ \xrightarrow[]{\text{BTNT C}}n_{CO_2}=0,12.2+0,3=0,54\left(mol\right)$So sánh: `0,54 < 0,72 => CO` dưPT rút gọn: $CO+O\rightarrow CO_2$ 0,54<-0,54CTHH: `Fe_xO_y``-> x : y = 0,36 : 0,54 = 2 : 3`CTHH: `Fe_2O_3`

๖ۣۜDũ๖ۣۜN๖ۣۜG · Answer

$n_{H_2}=\dfrac{8,064}{22,4}=0,36\left(mol\right)$PTHH: 2R + 2nHCl --> 2RCln + nH2 $\dfrac{0,72}{n}$<-----------------0,36$m_{tăng}=m_R-m_{H_2}=\dfrac{0,72}{n}.M_R-0,36.2=19,44\left(g\right)$=> $M_R=28n\left(g/mol\right)$Xét n = 2 thỏa mãn => MR = 56 (g/mol)=> R là Fe$n_{Fe}=0,36\left(mol\right)$$n_{CO\left(bđ\right)}=\dfrac{16,128}{22,4}=0,72\left(mol\right)$$n_{Ba\left(OH\right)_2}=0,21.2=0,42\left(mol\right)$; $n_{BaCO_3}=\dfrac{59,1}{197}=0,3\left(mol\right)$TH1: Kết tủa không bị hòa tanPTHH: Ba(OH)2 + CO2 --> BaCO3 + H2O 0,3<----0,3 FexOy + yCO --to--> xFe + yCO2 $\dfrac{0,3x}{y}$<-0,3=> $\dfrac{0,3x}{y}=0,36\Rightarrow\dfrac{x}{y}=\dfrac{6}{5}\left(L\right)$TH2: Kết tủa bị hòa tan 1 phầnPTHH: Ba(OH)2 + CO2 --> BaCO3 + H2O 0,42---->0,42---->0,42 BaCO3 + CO2 + H2O --> Ba(HCO3)2 0,12-->0,12=> $n_{CO_2}=0,42+0,12=0,54\left(mol\right)$PTHH: FexOy + yCO --to--> xFe + yCO2 $\dfrac{0,54x}{y}$<-0,54=> $\dfrac{0,54x}{y}=0,36\Rightarrow\dfrac{x}{y}=\dfrac{2}{3}$=> CTHH: Fe2O3Câu trả lời: $n_{H_2}=\dfrac{8,064}{22,4}=0,36\left(mol\right)$PTHH: 2R + 2nHCl --> 2RCln + nH2 $\dfrac{0,72}{n}$<-----------------0,36$m_{tăng}=m_R-m_{H_2}=\dfrac{0,72}{n}.M_R-0,36.2=19,44\left(g\right)$=> $M_R=28n\left(g/mol\right)$Xét n = 2 thỏa mãn => MR = 56 (g/mol)=> R là Fe$n_{Fe}=0,36\left(mol\right)$$n_{CO\left(bđ\right)}=\dfrac{16,128}{22,4}=0,72\left(mol\right)$$n_{Ba\left(OH\right)_2}=0,21.2=0,42\left(mol\right)$; $n_{BaCO_3}=\dfrac{59,1}{197}=0,3\left(mol\right)$TH1: Kết tủa không bị hòa tanPTHH: Ba(OH)2 + CO2 --> BaCO3 + H2O 0,3<----0,3 FexOy + yCO --to--> xFe + yCO2 $\dfrac{0,3x}{y}$<-0,3=> $\dfrac{0,3x}{y}=0,36\Rightarrow\dfrac{x}{y}=\dfrac{6}{5}\left(L\right)$TH2: Kết tủa bị hòa tan 1 phầnPTHH: Ba(OH)2 + CO2 --> BaCO3 + H2O 0,42---->0,42---->0,42 BaCO3 + CO2 + H2O --> Ba(HCO3)2 0,12-->0,12=> $n_{CO_2}=0,42+0,12=0,54\left(mol\right)$PTHH: FexOy + yCO --to--> xFe + yCO2 $\dfrac{0,54x}{y}$<-0,54=> $\dfrac{0,54x}{y}=0,36\Rightarrow\dfrac{x}{y}=\dfrac{2}{3}$=> CTHH: Fe2O3

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