a, m_CaO = 0,5.56 = 28 (g)

b, \(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

c, \(n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\)

d, \(V_{hhk}=0,2.22,4+0,3.22,4=11,2\left(l\right)\)

e, \(\%m_{Cu}=\dfrac{64}{64+32+16.4}.100\%=40\%\)

Bạn tham khảo nhé!

a) mCaO=nCaO.M(CaO)=0,5.56=28(g)

b) nCO2=V(CO2,dktc)=6,72/22.4=0,3(mol)

c) nH2SO4=mH2SO4/M(H2SO4)=24,5/98=0,25(mol)

d) V(hh H2,NH3)=(0,3+0,2).22,4=11,2(l)

e) %mCu/CuSO4=(64/160).100=40%

Chúc em học tốt!

a) m_{CuO = 0,25 x 80 = 20 (g)}

a)mCuO=0.25*(64+16)=20(g)

b)\(n_{MgCl_2}=\dfrac{19}{95}=0.2\left(mol\right)\)

Số phân từ MgCl₂ có trong 19g là 
0.2*6*10²³=1,2.10²³

c)
\(V_{hh}=\left(0.2+0.3+\dfrac{6.4}{32}\right).22,4=\left(0.5+0.2\right)=0.7\cdot22,4=15,68\left(l\right)\)

a) \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

\(n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\)

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(n_{Cu}=\dfrac{19,2}{64}=0,3\left(mol\right)\)

b) 

\(m_{CO_2}=44.0,5=22\left(g\right)\)

\(m_{H_2}=1,5.2=3\left(g\right)\)

\(m_{N_2}=2.28=56\left(g\right)\)

\(m_{CuO}=3.80=240\left(g\right)\)

c) \(n_{Cl_2}=\dfrac{14,2}{71}=0,2\left(mol\right)\)

\(n_{H_2}=\dfrac{4,8}{2}=2,4\left(mol\right)\)

\(n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\)

=> n_hh = 0,2 + 2,4 + 0,1 = 2,7 (mol)

=> V_hh = 2,7.22,4 = 60,48(l)

\(a.n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\\ n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ n_{Cu}=\dfrac{19,2}{64}=0,3\left(mol\right)\)

\(b.m_{CO_2}=0,5.44=22\left(g\right)\\ m_{H_2}=1,5.2=3\left(g\right)\\ m_{N_2}=2.28=56\left(g\right)\\ m_{CuO}=3.80=240\left(g\right)\)

\(c.n_{Cl_2}=\dfrac{14,2}{71}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{4,8}{2}=2,4\left(mol\right)\\ n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\\\Rightarrow n_{hh}=0,2+2,4+0,1=2,7\left(mol\right)\\ \Rightarrow V_{hh}=2,7.22,4=60,48\left(l\right)\)

\(a,n_{Mg}=\dfrac{2,4}{24}=0,1(mol); n_{Zn}=\dfrac{32,5}{65}=0,5(mol)\\ n_{Al}=\dfrac{2,7}{27}=0,1(mol); n_{Cu}=\dfrac{19,2}{64}=0,3(mol)\)

\(b,m_{CO_2}=0,5.44=22(g);m_{H_2}=1,5.2=3(g)\\ m_{N_2}=2.28=56(g);m_{CuO}=3.80=240(g)\)

\(c,n_{hh}=n_{Cl_2}+n_{H_2}+n_{O_2}=\dfrac{14,2}{71}+\dfrac{4,8}{2}+\dfrac{3,2}{32}=0,2+2,4+0,1=2,7(mol)\\ V_{hh}=2,7.22,4=60,48(l)\)

a, PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

\(2H_2+O_2\underrightarrow{t^o}2H_2O\)

Ta có: \(n_{CH_4}+n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\left(1\right)\)

Theo PT: \(n_{H_2O}=2n_{CH_4}+n_{H_2}=\dfrac{16,2}{18}=0,9\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,4\left(mol\right)\\n_{H_2}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,4.22,4}{11,2}.100\%=80\%\\\%V_{H_2}=20\%\end{matrix}\right.\)

b, Theo PT: \(n_{CO_2}=n_{CH_4}=0,4\left(mol\right)\Rightarrow V_{CO_2}=0,4.22,4=8,96\left(l\right)\)

n CO 2  = 0,4mol

Thể tích của khí CO 2 :   V CO 2  = 0,4 x 22,4 = 8,96l

a)

b) VCO2 = 22,4 .0,175 = 3,92l.

VH2 = 22,4 .1,25 = 28l.

VN2 = 22,4.3 = 67,2l.

c) Số mol của hỗn hợp khí bằng tổng số mol của từng khí.

nhh = nCO2 + nH2 + nN2 = 0,01 + 0,02 + 0,02 = 0,05 mol

Vhh khí = (0,01 + 0,02 + 0,02) . 22,4 = 1,12l.