\(\begin{array}{l} *SO_3:\\ \%S=\dfrac{32}{32+16\times 3}\times 100\%=40\%\\ \%O=\dfrac{16\times 3}{32+16\times 3}\times 100\%=60\%\\ *Fe_2O_3:\\ \%Fe=\dfrac{56\times 2}{56\times 2+16\times 3}\times 100\%=70\%\\ \%O=\dfrac{16\times 3}{56\times 2+16\times 3}\times 100\%=30\%\\ *CO_2:\\ \%C=\dfrac{12}{12+16\times 2}\times 100\%=27,27\%\\ \%O=\dfrac{16\times 2}{12+16\times 2}\times 100\%=72,73\%\end{array}\)

\(\begin{array}{l} PTHH:2Mg+O_2\xrightarrow{t^o} 2MgO\\ n_{Mg}=\dfrac{7,2}{24}=0,3\ (mol)\\ Theo\ pt:\ n_{O_2}=\dfrac{1}{2}n_{Mg}=0,15\ (mol)\\ \Rightarrow V_{O_2}=0,15\times 22,4=3,36\ (l)\\ Theo\ pt:\ n_{MgO}=n_{Mg}=0,3\ (mol)\\ \Rightarrow m_{MgO}=0,3\times 40=12\ (g)\end{array}\)

\(\begin{array}{l} a,\\ n_{K_2SO_4}=\dfrac{17,4}{174}=0,1\ (mol)\\ PTHH:K_2SO_4+BaCl_2\to BaSO_4\downarrow+2KCl\\ Theo\ pt:\ n_{BaSO_4}=n_{K_2SO_4}=0,1\ (mol)\\ \Rightarrow m_{BaSO_4}=0,1\times 233=23,3\ (g)\\ b,\\ Theo\ pt:\ n_{BaCl_2}=n_{K_2SO_4}=0,1\ (mol)\\ \Rightarrow m_{\text{dd BaCl_2}}=\dfrac{0,1\times 208}{10\%}=208\ (g)\\ m_{\text{dd spư}}=m_{K_2SO_4}+m_{\text{dd BaCl_2}}-m_{BaSO_4}\\ \Rightarrow m_{\text{dd spư}}=17,4+208-23,3=202,1\ (g)\\ Theo\ pt:\ n_{KCl}=2n_{K_2SO_4}=0,2\ (mol)\\ \Rightarrow C\%_{\text{dd spư}}=C\%_{KCl}=\dfrac{0,2\times 74,5}{202,1}\times 100\%=7,37\%\end{array}\)

\(\begin{array}{l} \text{Gọi kim loại kiềm là R.}\\ PTHH:2R+2H_2O\to 2ROH+H_2\uparrow\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\ (mol)\\ Theo\ pt:\ n_{R}=2n_{H_2}=0,4\ (mol)\\ \Rightarrow M_{R}=\dfrac{15,6}{0,4}=39\ (g/mol)\\ \Rightarrow \text{Kim loại kiềm cần tìm là Kali (K)}\end{array}\)

\(\begin{array}{l} a,\ PTHH:2Zn+O_2\xrightarrow{t^o} 2ZnO\\ b,\\ n_{Zn}=\dfrac{19,5}{65}=0,3\ (mol)\\ Theo\ pt:\ n_{O_2}=\dfrac{1}{2}n_{Zn}=0,15\ (mol)\\ \Rightarrow V_{O_2}=0,15\times 22,4=3,36\ (l)\\ c,\\ Theo\ pt:\ n_{ZnO}=n_{Zn}=0,3\ (mol)\\ \Rightarrow m_{ZnO}=0,3\times 81=24,3\ (g)\end{array}\)

                                                                                                   Giải

An kém Bình số viên bi là:

            10-7=3(viên bi)                                                                                                                                                                                                           Đ/S:3 viên bi