Bài 45: Hóa học và vấn đề môi trường

$n_{Mg_2P_2O_7} = \dfrac{0,235}{222} = 0,001(mol)$

$\Rightarrow n_P = 2n_{Mg_2P_2O_7} = 0,002(mol)$
$\%P = \dfrac{0,002.31}{1}.100\% = 6,2\%$

\(n_{H_2}=0,015\left(mol\right)\)

\(n_{H_2SO_4}=0,0152\left(mol\right)\)

\(n_{CuSO_4}=0,0152\left(mol\right)\)

\(Ba^{2+}+SO_4^{2-}-->BaSO_4\)

\(Cu^{2+}+2OH^--->Cu\left(OH\right)_2\)

a) Do 2 kim loại có cùng số mol \(\Rightarrow\) đặt \(n_{Na}=n_{Ba}=a\left(mol\right)\)

BT e : \(n_{Na}+2n_{Ba}=2n_{H_2}\Leftrightarrow a+2a=0,15.2\Rightarrow a=0,1\left(mol\right)\)

\(n_{Na^+}=n_{Na}=0,1\left(mol\right);n_{Ba^{2+}}=n_{Ba}=0,1\left(mol\right)\)

\(\Sigma n_{SO_4^{2-}}=n_{H_2SO_4}+n_{CuSO_4}=0,0152+0,0152=0,0304\left(mol\right)\)

\(\Rightarrow SO_4^{2-}pư\) hết \(\Rightarrow n_{BaSO_4}=0,0304.233=7,0832\left(g\right)\)

\(\Sigma n_{OH^-}=n_{Na^+}+2n_{Ba^{2+}}=0,1+0,1.2=0,3\left(mol\right)\)

\(n_{OH\left(Cu\left(OH\right)_2\right)}=n_{OH^-Bđ}-n_{Ba^{2+}dư}-n_{Na^+}=0,3-\left(0,1-0,0304\right)-0,1=0,1304\left(mol\right)\)

\(n_{Cu\left(OH\right)_2}=\dfrac{0,1304}{2}=0,0652\left(mol\right)\Rightarrow m_{Cu\left(OH\right)_2}=0,0652.98=6,3896\left(g\right)\)

\(\Rightarrow m=m_{BaSO_4}+m_{Cu\left(OH\right)_2}\)