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a) \(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\)
\(n_{CaCO_3}=\dfrac{25}{100}=0,25\left(mol\right)\)
\(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\)
\(n_{...}=\dfrac{1,5.10^{23}}{6.10^{23}}=0,25\left(mol\right)\)
b)
\(m_{ZnSO_4}=0,25.161=40,25\left(g\right)\)
\(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
\(m_{Cu}=0,3.64=19,2\left(g\right)\)
\(m_{Fe_2\left(SO_4\right)_3}=0,35.400=140\left(g\right)\)
d) \(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)
\(V_{Cl_2}=0,15.22,4=3,36\left(l\right)\)
\(V_{SO_2}=0,3.22,4=6,72\left(l\right)\)
a) \(m_{Cu}=0,6.64=38,4\left(g\right)\)
b) \(m_{\left(NH4\right)2SO4}=0,8.132=105,6\left(g\right)\)
300ml = 0,3l
\(n_{CuSO4}=0,2.0,3=0,06\left(mol\right)\)
Chúc bạn học tốt
a)
\(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\)
\(n_{Ca}=\dfrac{20}{40}=0,5\left(mol\right)\)
\(n_{CaCO_3}=\dfrac{25}{100}=0,25\left(mol\right)\)
\(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\)
\(n_{H_2O}=\dfrac{1,5.10^{23}}{6.10^{23}}=0,25\left(mol\right)\)
b)
\(m_{ZnSO_4}=0,25.161=40,25\left(g\right)\)
\(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
\(m_{Cu}=0,3.64=19,2\left(g\right)\)
\(m_{Fe_2\left(SO_4\right)_3}=0,35.400=140\left(g\right)\)
c)
\(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)
\(V_{Cl_2}=0,15.22,4=3,36\left(l\right)\)
\(V_{SO_2}=0,3.22,4=6,72\left(l\right)\)
\(V_{CH_4}=0,5.22,4=11,2\left(l\right)\)
a: \(n_{Fe}=\dfrac{14}{56}=0.25\left(mol\right)\)
\(n_{Ca}=\dfrac{20}{40}=0.5\left(mol\right)\)
\(PTK_{CO_2}=NTK_C+2NTK_O=12+2.16=44\left(\text{đ}.v.C\right)\\ PTK_{H_2SO_4}=NTK_H.2+NTK_S+4.NTK_O=2.1+32+4.16=98\left(\text{đ}.v.C\right)\)
Tương tự tính câu 3,4,5
\(1,M_{CO_2}=12+16\cdot2=44\left(g/mol\right)\\ 2,M_{H_2SO_4}=2+32+16\cdot4=98\left(g/mol\right)\\ 3,M_{Cu\left(NO_3\right)_2}=64+\left(14+16\cdot3\right)\cdot2=188\left(g/mol\right)\\ 4,M_{Fe_2\left(SO_4\right)_3}=56\cdot2+\left(32+16\cdot4\right)\cdot3=400\left(g/mol\right)\\ 5,M_{Na_3PO_4}=23\cdot3+31+16\cdot4=164\left(g/mol\right)\)
\(m_{Fe_2O_3}=16\cdot75\%=12\left(g\right)\)
\(n_{Fe_2O_3}=\dfrac{12}{160}=0.075\left(mol\right)\)
\(n_{CuO}=16\cdot25\%=4\left(g\right)\)
\(n_{CuO}=\dfrac{4}{80}=0.05\left(mol\right)\)
\(Fe_2O_3+3H_2\underrightarrow{^{^{t^0}}}2Fe+3H_2O\)
\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)
\(n_{H_2}=3\cdot0.075+0.05=0.275\left(mol\right)\)
a,\(m_{Fe_2O_3}=16.75\%=12\left(g\right)\Rightarrow n_{Fe_2O_3}=\dfrac{12}{160}=0,075\left(mol\right)\)
\(m_{CuO}=16-12=4\left(g\right)\Rightarrow n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)
PTHH: Fe2O3 + 3H2 → 2Fe + 3H2O
Mol: 0,075 0,225 0,15
PTHH: CuO + H2 → Cu + H2O
Mol: 0,05 0,05 0,05
\(\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right);m_{Cu}=0,05.64=3,2\left(g\right)\)
b,\(n_{H_2}=0,225+0,05=0,275\left(mol\right)\)
nCuSO4=\(\dfrac{16}{160}=0,1mol\)