Đáp án:
1A
2A
3A
5A
6D
7A
8C
9B
10A
11D
12A
13A
14B
15B
16D
17A
18D
19A
20C
21D
22C
23A
24D
25D
Giải thích các bước giải:
Câu 5
Gọi số mol Na2CO3,NaHCO3 lần lượt là x,y
$Na2CO3+H2SO4→Na2SO4+H2O+CO2$
$2NaHCO3+H2SO4→Na2SO4+2H2O+2CO2$
$n_{CO2}=0,56/22,4=0,025(mol)$
Ta có HPT sau
$\left \{ {{106x+84y=2,364} \atop {x+y=0,025}} \right.$ ⇒$\left \{ {{x=0,012} \atop {y=0,013}} \right.$
⇒%$m_{Na2CO3}=0,012.106/2,364.100$%$=53,81$%
Câu 7
$C2H4+Br2→C2H4Br2$
Ta có
$n_{Br2}2=4/160=0,025(mol)$
$nC2H4=nBr2=0,025(mol)$
⇒$V_{C2H4}=0,025.22,4=0,56(l)$
⇒%$V_{C2H4}=0,56/2,8.100$%$=20$%
Câu 9
$CaCO3→CaO+CO2$
$nCaCO3=100/100=1(mol)$
$nCO2=nCaCO3=1(mol)$
⇒$mCO2=1.44=44(g)$
⇒$H=33/44.100$%$=75$%
Câu 12
$C2H2+2Br2→C2H2Br4$
$nBr2=2nC2H2=0,2(mol)$
→$mBr2=0,2.160=32(g)$
câu 13
$nCO2=13,2/44=0,3(mol)$
→$nC=0,3(mol)$→$mC=0,3.12=3,6(g)$
→$mH=4,4-3,6=0,8(mol)$→$nH=0,8(mol)$
$n_C:n_H=0,3:0,8=3:8$
câu 15
$MnO2+4HCl→MnCl2+Cl2+2H2O$
$nMnO2=69,6/87=0,8(mol)$
$n_{Cl2}=nMnO2=0,8(mol)$
$V_{Cl2}=0,8.22,4=17,92(l)$
câu 16
$2C2H2+5O2→4CO2+2H2O$
$nO2=5/2nC2H2=2,5(mol)$
⇒$V_{O2}=2,5.22,4=56(l)$
⇒$V_{kk}=5V_{O2}=56.5=280(l)$
câu 19
$C2H2+2Br2→C2H2Br4$
$nBr2=5,6/160=0,0375(mol)$
$nC2H2=1/2nBr2=0,0175(mol)$
⇒$V_{C2H2}=0,0175.22,4=0,392(l)$
⇒%$V_{C2H2}=0,392/0,56.100$%$=70$%
câu 21
$2CH3COOH+Ca(OH)2→(CH3COO)2Ca+2H2O$
$n_{CH3COOH}=0,1.0,1=0,01(mol)$
$n_{Ca(OH)2}=1/2nCH3COOH=0,005(mol)$
⇒$CM_{Ca(OH)2}=0,005/0,1=0,05(l)=50ml$
câu 22
$Cl2+2NaOH→NaClO+NaCl+H2O$
$nCl2=1,12/22,4=0,05(mol)$
$nNaOH=2nCl2=0,1(mol)$
⇒$V_{NaOH}=0,1/1=0,1(l)$
câu 23
Gọi số mol mol CuO va FeO lần lượt la x,y
$CuO+CO→Cu+CO2$
$FeO+CO→Fe+CO2$
$nCO=0,84/28=0,03(mol)$
Ta có HPT
$\left \{ {{80x+72y=2,32} \atop {x+y=0,03}} \right.$⇒$\left \{ {{x=0,02} \atop {y=0,01}} \right.$
⇒$mCuO=0,02.80=1,6(g)$