Câu 1 :
a)
nCO2 = 13.2/44 = 0.3(mol)
VCO2 = 0.3*22.4 = 6.72 (l)
b)
nC4H10 = 8.96/22.4 = 0.4 (mol)
mC4H10 = 0.4*58 = 23.2 (g)
c)
nCaO = 3*10^23 / 6 *10^23 = 0.5 (mol)
nCa(OH)2 = 1.8*10^23 / 6*10^23 = 0.3 (mol)
mA = 0.5*56 + 0.3*74 = 50.2 (g)
Câu 1::
a) nCO2=13,2/44=0,3(mol)
=>V(CO2,đktc)=0,3.22,4=6,72(l)
b) nC4H10=8,96/22,4=0,4(mol)
->mC4H10=0,4.58= 23,2(g)
c) nCaO= (3.1023)/(6.1023)= 0,5(mol)
nCa(OH)2= (1,8.1023)/(6.1023)=0,3(mol)
=>mhhA= mCaO+ mCa(OH)2= 0,5.56 + 0,3.74= 50,2(g)
Câu 2 :
nX = 13.44/22.4 = 0.6 (mol)
MX = 32.4/0.6 = 54 (g/mol)
=> 4X + 6 = 54
=> X = 12
X là : Cacbon
Câu 1:
a,\(n_{CO_2}=\dfrac{13,2}{44}=0,3\left(mol\right)\Rightarrow V_{CO_2}=0,3.22,4=6,72\left(l\right)\)
b,\(n_{C_4H_{10}}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\Rightarrow m_{C_4H_{10}}=0,4.58=23,2\left(g\right)\)
c,\(n_{CaO}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\Rightarrow m_{CaO}=0,5.56=28\left(g\right)\)
\(n_{Ca\left(OH\right)_2}=\dfrac{1,8.10^{23}}{6.10^{23}}=0,3\left(mol\right)\Rightarrow m_{Ca\left(OH\right)_2}=0,3.74=22,2\left(g\right)\)
⇒ mhhA = 28+22,2=50,2 (g)