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a) \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)
b) \(n_{Al}=\dfrac{13,5}{27}=0,5\left(mol\right)\)
c) \(n_{CO_2}=\dfrac{11}{44}=0,25\left(mol\right)\)
d) \(m_{O_2}=\dfrac{4,958.0,99}{0,082.\left(273+25\right)}=0,2\left(mol\right)\)
e) \(m_{CH_4}=\dfrac{12,359.0,99}{0,082\left(273+25\right)}=0,5\left(mol\right)\)
a: \(n=\dfrac{28}{56}=0.5\left(mol\right)\)
b: \(n=\dfrac{13.5}{27}=0.5\left(mol\right)\)
\(a.\)
- \(n_{Fe}=\frac{11,2}{56}=0,2\left(mol\right)\)
- \(n_{H2SO4}=\frac{19,6}{98}=0,2\left(mol\right)\)
\(b.\)
- \(n_{SO2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow m_{SO2}=0,25\times64=16\left(gam\right)\)
- \(n_{H2}=\frac{22,4}{22,4}=1\left(mol\right)\)
\(\Rightarrow m_{H2}=1\times2=2\left(gam\right)\)
a) \(n_{Fe}=\frac{m}{M}=\frac{11,2}{56}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=\frac{m}{M}=\frac{19,6}{98}=0,2\left(mol\right)\)
b) \(n_{SO_2}=\frac{V}{22,4}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow m_{SO_2}=M.n=64.0,25=16\left(g\right)\)
* \(n_{H_2}=\frac{V}{22,4}=\frac{22,4}{22,4}=1\left(mol\right)\)
\(\Rightarrow m_{H_{ }_2}=M.n=2.1=2\left(g\right)\)
a) *mAl=n.M=0,9.27=24,3 (g)
* VAl=m/D=24,3/2,7=9cm3=0,009 lít
b) * mCl2=n.M=1,25.71=88,75 (g)
* VCl2=n.22,4=1,25.22,4=28 lít
c) * mNH3=n.M=0,86.17=14,62 (g)
*VNH3=n.22,4=0,86.22,4=19,264 (lít)
huyền trân Mình sửa lại nha SORRY
a) \(\left\{{}\begin{matrix}m_{Al}=27.0,9=24,3\left(g\right)\\V_{Al}=\dfrac{24,3}{2,7}=9\left(cm^3\right)=0,009\left(l\right)\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}m_{Cl2}=71.1,25=88,75\left(g\right)\\V_{Cl2}=22,4.1,25=28\left(g\right)\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}m_{NH3}=17.0,86=14,62\left(mol\right)\\V_{NH3}=22,4.0,86=19,264\left(l\right)\end{matrix}\right.\)
n của h2=1.2.1023:6.1023=0.2 mol
nSo2=6,4:64=0.1 mol
a,Vhh=[1,5+2,5+0.2+0,1] .22,4=96,32l
mhh=(1,5.32)+(2,5.28)+(0,2.2)+6,4=124,8g
nSO2 = 6,4 / 64 = 0,1 mol
nH2 = \(\frac{1,2\times10^{23}}{6\times10^{23}}=0,2\left(mol\right)\)
a/ Vhỗn hợp khí(đktc) = ( 0,1 + 0,2 + 1,5 + 2,5 ) x 22,4 = 96,32 lít
b/ mO2 = 1,5 x 32 = 48 gam
nN2 = 2,5 x 28 = 70 gam
nH2 = 0,2 x 2 = 0,4 gam
=> mhỗn hợp khí = 48 + 70 + 0,4 + 6,4 = 124,8 gam
\(nNH_3=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)
\(n=\dfrac{4.958}{22.4}=0,22\left(mol\right)\)