\(a)7437ml=7,437l\\ n_{N_2}=\dfrac{7,437}{24,79}=0,3mol\\ m_{N_2}=0,3.28=8,4g\\ b)n_{Cl_2}=\dfrac{2,479}{24,79}=0,1mol\\ m_{Cl_2}=0,1.71=7,1g\)

b, 14,9 g

a) đề bài??

b)

\(n_{SO_3\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ m_{SO_3}=n\cdot M=0,1\cdot\left(32+16\cdot3\right)=8\left(g\right)\)

\(n_{NH_3}=\dfrac{4,958}{24,79}=0,2(mol)\)

\(nNH_3=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

\(n=\dfrac{4.958}{22.4}=0,22\left(mol\right)\)

a) \(m_{Na2SO4}=0,2.142=28,4\left(g\right)\)

b) \(n_{H2\left(dktc\right)}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

\(m_{H2}=0,5.2=1\left(g\right)\)

 Chúc bạn học tốt

\(n_{O_2}=\dfrac{13,44}{22,4}=0,6mol\)

\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)

    0,4                                               0,6  ( mol )

\(m_{KClO_3}=0,4.122,5=49g\)

a) \(n_{Cl_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: 2Fe + 3Cl₂ --t^o--> 2FeCl₃

_____\(\dfrac{2}{15}\)<--0,2------------->\(\dfrac{2}{15}\)

=> m_Fe = \(\dfrac{2}{15}.56=7,467\left(g\right)\)

b) \(m_{FeCl_3}=\dfrac{2}{15}.162,5=21,667\left(g\right)\)

Ta có: \(n_{C_2H_2}=\dfrac{0,336}{22,4}=0,015\left(mol\right)\)

Bảo toàn nguyên tố Cacbon: \(n_{CO_2}=2n_{C_2H_2}=0,03\left(mol\right)\)

\(\Rightarrow m_{CO_2}=0,03\cdot44=1,32\left(g\right)\)

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