Đáp án:

Giải thích các bước giải:

\(\text{Bài 4:}\\ a,\ PTHH:\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2↑\\ b,\ n_{Al}=\dfrac{4,05}{27}=0,15\ mol.\\ n_{H_2}=\dfrac{3,36}{22,4}=0,15\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,15}{2}>\dfrac{0,15}{3}\\ ⇒Al\ dư.\\ Theo\ pt:\ n_{Al}(pư)=\dfrac{2}{3}n_{H_2}=0,1\ mol.\\ ⇒m_{Al}=0,1\times 27=2,7\ g.\\ c,\ Theo\ pt:\ n_{Al_2(SO_4)_3}=\dfrac{2}{3}n_{H_2}=0,1\ mol.\\ ⇒m_{Al_2(SO_4)_3}=0,1\times 342=34,2\ g.\\ \text{Bài 5:}\\ a,\ PTHH:4P+5O_2\xrightarrow{t^o} 2P_2O_5\\ b,\ n_{O_2}=\dfrac{5,6}{22,4}=0,25\ mol.\\ n_{P_2O_5}=\dfrac{10,65}{31}=0,075\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,25}{5}>\dfrac{0,075}{2}\\ ⇒O_2\ dư.\\ Theo\ pt:\ n_{P}=2n_{P_2O_5}=0,15\ mol.\\ ⇒m_{P}=0,15\times 31=4,65\ g.\\ Theo\ pt:\ n_{O_2}=\dfrac{5}{2}n_{P_2O_5}=0,1875\ mol.\\ ⇒m_{O_2}=0,1875\times 32=6\ g.\\ c,\ n_{O_2}(dư)=0,25-0,1875=0,0625\ mol.\\ ⇒m_{O_2}(dư)=0,0625\times 32=2\ g.\\ \text{Bài 6:}\\ a,\ PTHH:3Fe+2O_2\xrightarrow{t^o} Fe_3O_4\\ b,\ n_{O_2}=\dfrac{8,4}{22,4}=0,375\ mol.\\ n_{Fe_3O_4}=\dfrac{34,8}{232}=0,15\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,375}{2}>\dfrac{0,15}{1}\\ ⇒O_2\ dư.\\ Theo\ pt:\ n_{Fe}=3n_{Fe_3O_4}=0,45\ mol.\\ ⇒m_{Fe}=0,45\times 56=25,2\ g.\\ c,\ Theo\ pt:\ n_{O_2}=2n_{Fe_3O_4}=0,3\ mol.\\ ⇒V_{O_2}=0,3\times 22,4=6,72\ lít.\\ d,\ PTHH:2KClO_3\xrightarrow{t^o} 2KCl+3O_2\\ Theo\ pt:\ n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,25\ mol.\\ ⇒m_{KClO_3}=0,25\times 122,5=30,625\ g.\\ \text{Bài 7:}\\ a,\ PTHH:2NaOH+CO_2\to Na_2CO_3+H_2O\\ b,\ n_{CO_2}=\dfrac{2,24}{22,4}=0,1\ mol.\\ Theo\ pt:\ n_{NaOH}=2n_{CO_2}=0,2\ mol.\\ \text{Đổi 200 ml = 0,2 lít.}\\ ⇒CM_{NaOH}=\dfrac{0,2}{0,2}=1\ M.\\ c,\ Theo\ pt:\ n_{Na_2CO_3}=n_{CO_2}=0,1\ mol.\\ ⇒m_{Na_2CO_3}=0,1\times 106=10,6\ g.\\ \text{Bài 8:}\\ a,\ PTHH:MnO_2+4HCl\to MnCl_2+Cl_2↑+2H_2O\\ b,\ n_{MnO_2}=\dfrac{8,7}{87}=0,1\ mol.\\ Theo\ pt:\ n_{HCl}=4n_{MnO_2}=0,4\ mol.\\ ⇒C\%_{HCl}=\dfrac{0,4\times 36,5}{116,8}\times 100\%=12,5\%\\ c,\ Theo\ pt:\ n_{Cl_2}=n_{MnO_2}=0,1\ mol.\\ ⇒m_{Cl_2}=0,1\times 71=7,1\ g.\\ d,\ Theo\ pt:\ n_{MnCl_2}=n_{MnO_2}=0,1\ mol.\\ m_{\text{dd spư}}=m-{MnO_2}+m_{\text{dd HCl}}-m_{Cl_2}\\ ⇒m_{\text{dd spư}}=8,7+116,8-7,1=118,4\ g.\\ ⇒C\%_{MnCl_2}=\dfrac{0,1\times 126}{118,4}\times 100\%=10,64\%\\ \text{Bài 9:}\\ a,\ PTHH:2NaOH+CuSO_4\to Na_2SO_4+Cu(OH)_2↓\\ b,\ \text{Đổi 400 ml = 0,4 lít.}\\ n_{CuSO_4}=0,4\times 0,225=0,09\ mol.\\ Theo\ pt:\ n_{Na_2SO_4}=n_{CuSO_4}=0,09\ mol.\\ V_{\text{dd spư}}=V_{\text{dd tpư}}=0,4\ lít.\\ ⇒CM_{Na_2SO_4}=\dfrac{0,09}{0,4}=0,225\ M.\\ \text{Bài 10:}\\ a,\ PTHH:FeS+2HCl\to FeCl_2+H_2S↑\\ b,\ n_{HCl}=\dfrac{73\times 10\%}{36,5}=0,2\ mol.\\ Theo\ pt:\ n_{H_2S}=\dfrac{1}{2}n_{HCl}=0,1\ mol.\\ ⇒m_{H_2S}=0,1\times 34=3,4\ g.\\ c,\ Theo\ pt:\ n_{FeS}=\dfrac{1}{2}n_{HCl}=0,1\ mol.\\ ⇒m_{\text{dd spư}}=m_{FeS}+m_{\text{dd HCl}}-m_{H_2S}\\ ⇒m_{\text{dd spư}}=0,1\times 88+73-3,4=78,4\ g.\\ d,\ Theo\ pt:\ n_{FeCl_2}=\dfrac{1}{2}n_{HCl}=0,1\ mol.\\ ⇒C\%_{FeCl_2}=\dfrac{0,1\times 127}{78,4}\times 100\%=16,2\%\)

chúc bạn học tốt!