a) \(n_{Fe_2O_3}=\frac{32}{160}=0,2\left(mol\right)\)

PTHH : \(Fe_2O_3+3H_2-t^o->2Fe+3H_2O\)

Theo pthh : \(n_{H_2}=3n_{Fe_2O_3}=0,6\left(mol\right)\)

=> \(V_{H_2}=0,6\cdot22,4=13,44\left(l\right)\)

b) Theo pthh : \(n_{H_2O}=n_{H_2}=0,6\left(mol\right)\)

=> \(m_{H_2O}=0,6\cdot18=10,8\left(g\right)\)

c) Theo pthh : \(n_{Fe}=2n_{Fe_2O_3}=0,4\left(mol\right)\)

=> \(m_{Fe}=0,4\cdot56=22,4\left(g\right)\)

1 we learn how things work in physics?

2 why don't we go to the movies tonight?

3 i am interesting in collecting stamps very much

1.We understand how things work thanks to physics

We learn  how things work thanks to where

2.Shall we go to the movies tonight? 

Why don't we go to the movie tonight ?

3.I to collect stamps very much

I am intrested in collecting stamps

4.our summer vacation often lasts for almost 3 months

We often have almost 3 months for our summer vacation 

a. \(n_{HCl}=\frac{44,8}{22,4}=2mol\)

\(\rightarrow m_{HCl}=2.36,5=73g\)

\(\rightarrow C\%_{HCl}=\frac{73.100}{73+327}=18,25\%\)

b. \(n_{CaCO_3}=\frac{50}{100}=0,5mol\)

\(n_{HCl}=\frac{250.18,25\%}{36,25}=1,25mol\)

PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

Từ phương trình \(n_{CaCl_2}=n_{CO_2}=0,5mol\)

\(n_{HCl\text{pứ}}=0,5.2=1mol\)

\(\rightarrow n_{HCl\text{dư}}=1,25-1=0,25mol\)

\(n_{ddsaupu}=50+250-0,5.44=278g\)

\(\rightarrow C\%_{CaCl_2}=\frac{0,5.111.100}{278}=19,96\%\)

\(\rightarrow C\%_{HCl}=\frac{0,25.36,5.100}{278}=3,28\%\)

giúp mk vs ạ

Gọi CTHH của A là : \(C_xH_y\)

Có : \(M_{C_xH_y}=28\cdot2=56\) (g/mol)

Lại có : \(\frac{12x}{56}\cdot100\%=85,7\%\Rightarrow x=4\)

                \(\frac{y}{56}\cdot100\%=14,3\%\Rightarrow y=8\)

Vậy CTHH của hợp chất A là C₄H₈

\(n_{C_3H_6O_2}=0,05\left(mol\right)\)

\(n_{CO_2}=3n_{C_3H_6O_2}=3.0,05=0,15\left(mol\right)\)

\(\Rightarrow V_{CO_2}=22,4.0,15=3,36\left(l\right)\)

\(n_{H_2O}=\frac{6}{2}n_{C_3H_6O_2}=0,15\left(mol\right)\)

\(\Rightarrow m_{H_2O}=18.0,15=2,7\left(g\right)\)

a. \(V_{N_2\left(ĐKTC\right)}=n_{N_2}.22,4=0,05.22,4=1,12l\)

b. \(4,4kg=4400g\)

\(\rightarrow n_{CO_2}=\frac{m_{CO_2}}{M_{CO_2}}=\frac{4400}{44}=100mol\)

\(\rightarrow V_{CO_2\left(ĐKTC\right)}=n_{CO_2}.22,4=100.22,4=2240l\)

c. \(n_{O_2}=\frac{A_{O_2}}{N}=\frac{1,2.10^{22}}{6.10^{23}}=0,02mol\)

\(\rightarrow m_{O_2}=n_{O_2}.M_{O_2}=0,02.32=0,64g\)

\(\rightarrow V_{O_2\left(ĐKTC\right)}=n_{O_2}.22,4=0,448l\)

Bảo toàn khối lượng

\(m_{O_2}=m_{Oxit}-m_{KL}=48,84-34,44=14,4g\)

\(\rightarrow n_{O_2}=\frac{14,4}{32}=0,45mol\)

BTNT (O) \(n_{H_2O}=2n_{O_2}=0,9mol\)

\(n_{H_2}=\frac{4,032}{22,4}=0,18mol\)

BTNT (H)  \(n_{HCl}=2n_{H_2O}+2n_{H_2}=2,16mol\)

\(\rightarrow m_{HCl}=2,16.36,5=78,84g\)

\(m_{H_2}=0,18.2=0,36g\) và \(m_{H_2O}=0,9.18=16,2g\)

Bảo toàn khối lượng \(m_A+m_{HCl}=m_{\text{muối}}+m_{H_2O}+m_{H_2}\)

\(\rightarrow m_{\text{muối}}=48,84+78,84-0,36-16,2=111,12g\)

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