Đáp án:
$a/$
$TH1:$
$\%mCu_2O=66,06\%$
$\%mMn_2O_7=33,94\%$
$TH2:$
$\%mCuO=73.4\%$
$\%mMnO_2=26,6\%$
$b/$
$VddHCl=0,1lit$
Giải thích các bước giải:
a/
$D \left \{ {{R_2O_x}\\M_2O_y\atop {Al}} \right. \buildrel{{t^o}}\over\longrightarrow (E)\left \{ {{Al_2O_3}\\Al _{dư}\\R \atop {M}} \right.$
TN1:
Ta có phản ứng sinh ra khí $⇒Al dư$
$nH_2=\frac{0,504}{22,4}=0,0225$
$Al^0 \to Al^{3+}+3e$
$2H^{+1} +2e\to H^0_2$
$BTe⇒3nAl=2nH_2⇔nAl_{dư}=0,015$
$nAl_{\text{ban đầu}}=0,015.2=0,03$
TN2:
$nNO=\frac{1,344}{22,4}=0,06$
Gọi $a$ là hóa trị của $M$
$N^{+5}+3e \to N^{+2}$
$M^0 \to M^{a+}+ae$
$BTe⇒anM=3nNO⇔nM=\frac{0,18}{a}$
$MM=\frac{5,76a}{0,18}=32a$
$+a=2⇒M=64(Cu)$
$nCu_{\text{ban đầu}}=\frac{5,76}{64}.2=0,18$
$nH_2=\frac{1,176}{22,4}=0,0525$
Gọi $b$ là hóa trị của $R$
$Al^0 \to Al^{3+}+3e$
$R^0 \to R^{b+}+be$
$2H^{+1} +2e\to H_2^0$
$BTe⇒3nAl+bnR=2nH_2$
$⇔nR=\frac{0,0525.2-0,015.3}{b}=\frac{0,06}{b}$
$nR_\text{ ban đầu}=\frac{0,12}{b}$
$mAl_{\text{trong E}}=25,83.24,042\%=6,21g$
$mAl_{\text{trong Al2O3}}=6,21-0,03.27=5,4g$
$nAl_{\text{trong Al2O3}}=\frac{5,4}{27}=0,2⇒nAl_2O_3=0,1$
$mR=25,83-mAl_2O_3-mAl-mCu=25,83-0,1.102-0,03.27-0,18.64=3,3g$
$MR=\frac{3,3b}{0,12}=27,5b$
$+b=2⇒R=55(Mn)$
$nMn=\frac{3,3}{55}=0,06$
$D \left \{ {{R_2O_x}\\M_2O_y\atop {Al}} \right. \buildrel{{t^o}}\over\longrightarrow (E)\left \{ {{Al_2O_3(0,1mol)}\\Al _{dư}(0,03mol)\\Mn(0,06mol) \atop {Cu(0,18mol)}} \right.$
Ta có $C:\left \{ {{Mn_2O_x} \atop {Cu_2Oy}} \right.$
$BTNT "Mn" ⇒2nMn_2O_x=nMn⇔nMn_2O_x=0,03$
$BTNT" Cu" ⇒2nCu_2O_y=nCu⇔nCu_2O_y=0,09$
$BTNT "O" ⇒xnMn_2O_x+ynCu_2O_y=3nAl_2O_3=0,3$
$⇔0,03x+0,09y=0,3$
$+y=1⇒x=7(Nhận)$
$⇒Cu_2O; Mn_2O_7$
$+y=2⇒x=4(Nhận)$
$⇒CuO; MnO_2$
$TH1: Cu_2O ;Mn_2O_7$
$BTNT" Cu" ⇒2nCu_2O=nCu⇔nCu_2O=0,09$
$BTNT "Mn" ⇒2nMn_2O_7=nMn⇔nMn_2O_7=0,03$
$\%mCu_2O=\frac{0,09.144}{0,09.144+0,03.222}.100=66,06\%$
$\%mMn_2O_7=100-66,06=33,94\%$
$TH2:CuO; MnO_2$
$BTNT" Cu" ⇒nCuO=nCu⇔nCuO=0,18$
$BTNT "Mn" ⇒nMnO=nMn⇔nMnO_2=0,06$
$\%mCuO=\frac{0,18.80}{0,18.80+0,06.87}.100=73.4\%$
$\%mMnO_2=100-73,4=26,6\%$
b/
Ta có $nO^{2-}=3nAl_2O_3=0,1,3=0,3$
$2H^++O^{2-} \to H_2O$
0,6 0,3
$nHCl=nH^+=0,6$
$VddHCl=\frac{0,6}{6}=0,1lit$
