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nCaCO3=10/100=0,1 mol
CaCO3 →CaO + CO2 (đk to)
0,1 0,1 0,1 mol
VCO2=0,1.22,4=2,24 l
mCaO=0,1.56=5,6 g
Bài 2 :
\(n_{CaCO3}=\dfrac{10}{100}=0,1\left(mol\right)\)
Pt : \(CaCO_3\underrightarrow{t^o}CaO+CO_2|\)
1 1 1
0,1 0,1 0,1
a) \(n_{CO2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(V_{CO2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)
b) \(n_{CaO}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
⇒ \(m_{CaO}=0,1.56=5,6\left(g\right)\)
Chúc bạn học tốt
Bài 1 :
\(n_{O2}=\dfrac{32}{32}=1\left(mol\right)\)
Pt : \(O_2+2Mg\underrightarrow{t^o}2MgO|\)
1 2 2
1 2 1
a) \(n_{Mg}=\dfrac{1.2}{1}=2\left(mol\right)\)
⇒ \(m_{Mg}=2.24=48\left(g\right)\)
c) \(n_{MgO}=\dfrac{2.1}{2}=1\left(mol\right)\)
⇒ \(m_{MgO}=1.40=40\left(g\right)\)
Chúc bạn học tốt
a, PTHH: CaCO3---> CaO+ CO2
Ta có nCaCO3=10/100=0,1mol
theo PTHH ta có nCO2=nCaCO3=0,1mol
=> VCO2=0,1.22,4=2,24 lít
b, Theo PTHH ta có : nCaO= nCaCO3=0,1mol
=> mCaO=0,1.56=5,6g
Chúc bạn học tốt ~
a) \(n_{CH_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,4---------------->0,4
=> \(V_{CO_2}=0,4.22,4=8,96\left(l\right)\)
b) \(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Xét tỉ lệ \(\dfrac{0,4}{1}>\dfrac{0,4}{2}\) => CH4 dư, O2 hết
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,4-------->0,2
=> \(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, Ta có: \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,2}{1}\), ta được H2 dư.
Theo PT: \(n_{H_2\left(pư\right)}=n_{Cu}=n_{H_2O}=n_{CuO}=0,15\left(mol\right)\)
\(\Rightarrow n_{H_2\left(dư\right)}=0,2-0,15=0,05\left(mol\right)\Rightarrow m_{H_2\left(dư\right)}=0,05.2=0,1\left(g\right)\)
\(m_{Cu}=0,15.64=9,6\left(g\right)\)
\(m_{H_2O}=0,15.18=2,7\left(g\right)\)
\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\\ n_{CaCO_3}=n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\m_{CaCO_3}=0,1.100=10\left(g\right)\)
Zn+H2SO4->ZnSO4+H2
0,11-------------------------0,11
2KMnO4-tO>K2MnO4+MnO2+O2
0,06-------------------------------------0,03
2H2+O2-to>2H2O
0,06---0,03-0,06
n Zn=0,11 mol
n KMnO4=0,06 mol
=>H2 du2
=>m H2O=0,06.18=1,08g
\(n_{Zn}=\dfrac{7,15}{65}=0,11\left(mol\right)\\
pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,11 0,11
\(n_{KMnO_4}=\dfrac{9,48}{158}=0,06\left(mol\right)\\
pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,06 0,03
\(pthh:2H_2+O_2\underrightarrow{t^o}2H_2O\\
LTL:\dfrac{0,11}{2}>\dfrac{0,03}{1}=>H_2d\text{ư}\)
theo pthh : nH2O =2 nO2= 0,12 (mol)
=> mH2O = 0,12 . 18 = 2,16(g)
PTPƯ:
\(CaCO_3\underrightarrow{t^o}CaO+CO_2\)\(\uparrow\)
0,015 0,015 0,015
\(nCaCO_3=\dfrac{1,5}{100}=0,015mol\)
\(mCaO=0,015.56=0,84\)(tấn)
\(n_{KMnO_4}=\dfrac{31,6}{158}=0,2mol\)
\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
0,2 0,1 0,1 0,1
a)\(V_{O_2}=0,1\cdot22,4=2,24l\)
b)\(m_{CRắn}=m_{K_2MnO_4}+m_{MnO_2}=0,1\cdot197+0,1\cdot87=28,4g\)
c)\(n_{CH_4}=\dfrac{11,2}{22,4}=0,5mol\)
\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
0,5 0,1 0 0
0,05 0,1 0,05 0,1
0,45 0 0,05 0,1
\(V_{CO_2}=0,05\cdot22,4=1,12l\)
\(m_{H_2O}=0,1\cdot18=1,8g\)
PTHH: 2KMnO4--to-> K2MnO4+MnO2+O2
0,2----------------0,1---------0,1-----0,1
b, nKMnO4= \(\dfrac{31,6}{158}\)=0,2 mol
Theo pt: nO2=\(\dfrac{1}{2}\).0,2=0,1 mol
=> VO2= 0,1.22,4= 2,24 l
=>m cr=0,1.197+0,1.87=28,4g
CH4+2O2-to>CO2+2H2O
0,5-----0,25-----0,5
n CH4=\(\dfrac{11,2}{22,4}\)=0,5 mol
=>Oxi du
=>V CO2=0,25.22,4=5,6l
=>m H2O=0,5.18=9g
$a) CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$
b) $n_{CH_4} = \dfrac{3,92}{22,4} = 0,175(mol)$
$n_{O_2} = \dfrac{3,84}{32} = 0,12(mol)$
Ta thấy : $n_{CH_4} : 1 > n_{O_2} : 2$ nên $CH_4$ dư
$n_{CH_4\ pư} = \dfrac{1}{2}n_{O_2} = 0,06(mol)$
$\Rightarrow m_{CH_4\ dư} = (0,175 - 0,06).16 = 1,84(gam)$
c) $2NaOH + CO_2 \to Na_2CO_3 + H_2O$
Theo PTHH :
$n_{Na_2CO_3} = n_{CO_2} = \dfrac{1}{2}n_{CH_4} = 0,06(mol)$
$m_{Na_2CO_3} = 0,06.106 = 6,36(gam)$
a) $\rm n_{CaCO_3} = \dfrac{10}{100} = 0,1 (mol)$
PTHH: $\rm CaCO_3 \xrightarrow{t^o} CaO + CO_2$
Theo PT: $\rm n_{CO_2} = n_{CaO} = n_{CaCO_3} = 0,1 (mol)$
$\rm \rightarrow V_{CO_2} = 0,1.22,4 = 2,24 (l)$
b) $\rm m_{CaO} = 0,1.56 = 5,6 (g)$