Ta có: \(n_{Al_4C_3}=\dfrac{72}{144}=0,5\left(mol\right)\)

Bảo toàn nguyên tố: \(\left\{{}\begin{matrix}n_{Al\left(lý.thuyết\right)}=4n_{Al_4C_3}=2\left(mol\right)\\n_{C\left(lý.thuyết\right)}=3n_{Al_4C_3}=1,5\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al\left(thực\right)}=\dfrac{2\cdot27}{60\%}=90\left(g\right)=b\\m_{C\left(thực\right)}=\dfrac{12\cdot1,5}{60\%}=30\left(g\right)=a\end{matrix}\right.\)

\(c_{hợp.kim}=30\%.380+70\%.130=205\left(\dfrac{j}{kg.K}\right)\)

PT: \(2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\)

_____x______\(\dfrac{3}{2}x\)___________________\(\dfrac{3}{2}x\) (mol)

Ta có: m _{thanh nhôm tăng} = m_Cu - m_Al

\(\Rightarrow11,5=\dfrac{3}{2}x.64-27x\Rightarrow x=\dfrac{1}{6}\left(mol\right)\)

\(\Rightarrow n_{CuSO_4}=0,25\left(mol\right)\)

Mà: n_MgSO4:n_CuSO4 = 3:2 ⇒ n_MgSO4 = 0,375 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgSO_4}=\dfrac{0,375.120}{0,375.120+0,25.160}.100\%\approx52,94\%\\\%m_{CuSO_4}\approx47,06\%\end{matrix}\right.\)

Tóm tắt:

\(m_1=2,5kg\)

\(t_1=20^oC\)

\(t_2=70^oC\)

\(c_1=880J/kg.K\)

==========

a) \(Q_1=?J\)

b) \(t_3=20^oC\)

\(t=40^oC\)

\(c_2=4200J/kg.K\)

c) \(t_1'=83^oC\)

\(m_1'=?kg\)

\(t_2'=25^oC\)

\(m_2'=?kg\)

a) Nhiệt lượng nhôm nhận được:

\(Q_1=m_1.c_1.\left(t_2-t_1\right)=2,5.880.\left(70-20\right)=110000J\)

b) Khối lượng của nước:

Theo pt cân bằng nhiệt:

\(Q_1=Q_2\)

\(\Leftrightarrow m_1.c_1.\left(t_2-t\right)=m_2.c_2.\left(t-t_3\right)\)

\(\Leftrightarrow m_2=\dfrac{m_1.c_1.\left(t_2-t\right)}{c_2.\left(t-t_3\right)}\)

\(\Leftrightarrow m_2=\dfrac{2,5.880.\left(70-40\right)}{4200.\left(40-20\right)}\approx0,8\left(kg\right)\)

c) Theo pt cân bằng nhiệt:

\(Q_1'=Q_2'\)

\(\Leftrightarrow m_1'.c_2.\left(t_1'-t\right)=m_2'.c_2.\left(t-t_2'\right)\)

\(\Leftrightarrow m_1'.\left(83-40\right)=m_2'.\left(40-25\right)\)

\(\Leftrightarrow43m_1'=15m_2'\left(k\right)\)

Mà ta có: \(m_1'+m_2'=m_{hh}\Rightarrow m_2'=m_{hh}-m_1'=0,8-m_1'\)

Thay vào k ta có:

\(43m_1'=15\left(0,8-m_1'\right)\)

\(\Leftrightarrow43m_1'=12-15m_1\)

\(\Leftrightarrow43m_1'+15m_1'=12\)

\(\Leftrightarrow58m_1'=12\)

\(\Leftrightarrow m_1'=\dfrac{12}{58}\approx0,21\left(kg\right)\)

Khối lượng nước ở 25^oC là:

\(m_2'=0,8-0,21=0,59\left(kg\right)\)

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right);n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(a,PTHH:4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)

\(b,Vì:\dfrac{0,1}{4}< \dfrac{0,2}{3}\Rightarrow O_2dư\\ c,n_{Al_2O_3}=\dfrac{n_{Al}}{2}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ \Rightarrow m_{Al_2O_3}=0,05.102=5,1\left(g\right)\\ d,n_{O_2\left(dư\right)}=\left(0,2-\dfrac{3}{4}.0,1\right)=0,125\left(mol\right)\\ \Rightarrow m_{O_2\left(dư\right)}=32.0,125=4\left(g\right)\)

đen thui lun ;-;