\(a)n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ 0,2\leftarrow-0,3\leftarrow-0,1\leftarrow---0,3\)

\(a=m_{Al}=0,2.27=5,4g\\ b)m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\\ c)C_{\%H_2SO_4}=\dfrac{0,3.98}{100}\cdot100=29,4\%\)

a)nH2​​=22,46,72​=0,3mol2Al+3H2​SO4​→Al2​(SO4​)3​+3H2​0,2←−0,3←−0,1←−−−0,3

�=���=0,2.27=5,4��)���2(��4)3=0,1.342=34,2��)�%�2��4=0,3.98100⋅100=29,4%a=mAl​=0,2.27=5,4gb)mAl2​(SO4​)3​​=0,1.342=34,2gc)C%H2​SO4​​=1000,3.98​⋅100=29,4%

\(a_1,m_{CaCO_3}=0,25.100=25(g)\\ a_2,m_{SO_2}=\dfrac{3,36}{22,4}.64=9,6(g)\\ a_3,m_{H_2SO_4}=\dfrac{9.10^{23}}{6.10^{23}}.98=147(g)\)

a) \(Pt:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b) \(n_{Fe}=\dfrac{0,56}{56}=0,01mol\)

Theo pt: \(n_{FeSO_4}=n_{Fe}=0,01mol\)

\(\Rightarrow m_{FeSO_4}=0,01.152=1,52g\)

Theo pt: \(n_{H_2}=n_{Fe}=0,01mol\)

\(\Rightarrow V_{H_2}=0,01.22,4=0,224lít\)

c) \(Theopt:nH_2SO_4=n_{Fe}=0,01mol\)

\(\Rightarrow m_{H_2SO_4}=0,01.98=0,98g\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,98.100}{19,6}=5g\)

a) n N2=\(\frac{0.54.10^{23}}{6.10^{23}}=0,09.10^{23}\)

b) n CO2=\(\frac{4,4}{44}=0,1\left(mol\right)\)

V CO2=0,1.22,4=2,24(l)

c) n N2O5=\(\frac{8,961}{22,4}=0,4\left(mol\right)\)

m N2O5=0,4.108=43,2(g)

d) m Fe(OH)3=1,5.107=160,5(g)

\(n_{H_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\\ m_{H_2}=1,5.2=3\left(g\right)\)

PTHH : 2Al + H₂SO₄ -> Al₂SO₄ + H₂

Theo ĐLBTKL

\(m_{Al}+m_{H_2SO_4}=m_{Al_2SO_4}+m_{H_2}\\ \Rightarrow m_{H_2SO_4}=\left(171+3\right)-2,7=171,3\left(g\right)\)

Em coi lại đề bài nhé! 0,1 mol nhôm không thể tạo ra được 0,5mol muối nhôm sunfat và 1,5 mol khí hidro

Các bạn dưới làm chắc không để ý đề bài!

pthh: 2Al+3H\(_2\)SO\(_4\)→Al\(_2\)(SO4)\(_3\)+3H\(_2\)↑

nH\(_2=33,6:22,4=1,5\left(mol\right)\)

\(mH_2=1,5.2=3\left(g\right)\)

\(nAl_2\left(SO_4\right)=171:150=1,14\left(mol\right)\)

\(mAl_2\left(SO_4\right)_3=1,14.342=389,88\left(g\right)\)

BTKL : mAl + mH\(_2\)SO\(_4\) = m Al\(_2\)(SO4)\(_3\)  + m H\(_2\)

           2,7    +  mH\(_2\)SO\(_4\) =  389,88       +   3

=> \(mH_2SO_4=\left(389,88+3\right)-2,7=390,18\left(g\right)\)  

\(Pt: 2Al+3H_2SO_4 \rightarrow Al_2(SO_4)_3 + 3H_2\)

\(a.n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo pt: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Al}=a=0,2.27=5,4\left(g\right)\)

\(b.n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\)

\(c.\)Theo pt: \(n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,3.98=29,4g\)

\(C_{\%}H_2SO_4=\dfrac{29,4}{100}.100\%=29,4\%\)