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

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

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

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

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

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 \(m_A+m_{HCl}=m_{\text{muối}}+m_{H_2O}+m_{H_2}\)

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

vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

giúp mình với chiều mình đi học r

\(X+O_2\rightarrow CO_2+H_2O\)

Đặt \(a\left(mol\right)=CO_2\)

\(N_{H_2O}=2N_{CO_2}=2a\left(mol\right)\)

BTKL: \(m_X+m_{O_2}=m_{CO_2}+m_{H_2O}\)

             \(16+64=44A+18.2a\)

\(\Leftrightarrow80a=80\)

\(\rightarrow a=1\)

\(m_{CO_2}=1.44=44g\)

\(m_{H_2O}=2.1.18=36g\)

a. \(Mg+CU\left(NO_3\right)_2\rightarrow Mg\left(NO_3\right)_2+Cu\downarrow\)

b. \(n_{Cu\left(\text{pứ}\right)}=a\left(mol\right)\)

\(\rightarrow m_{Mg\left(\tan\right)}=24a\) và \(m_{Cu\left(bám\right)}=64a\)

\(\rightarrow40-m_{Mg\left(\tan\right)}+m_{Cu\left(bám\right)}=48\)

\(\rightarrow40-24a+64a=48\)

\(\rightarrow a=0,2mol\)

c. \(m_{Cu}=0,2.64=12,8g\)