Answer 1

Ta có: \(n_{HCl}=0,3\cdot1=0,3\left(mol\right)\) \(\Rightarrow m_{HCl}=0,3\cdot36,5=10,95\left(g\right)\)

Bảo toàn nguyên tố: \(n_{H_2O}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\) \(\Rightarrow m_{H_2O}=0,15\cdot18=2,7\left(g\right)\)

Bảo toàn nguyên tố: \(m_{muối}=m_{oxit}+m_{HCl}-m_{H_2O}=12,71\left(g\right)\)

Bảo toàn nguyên tố: \(n_{NaOH}=0,3\cdot1=0,3\left(mol\right)=n_{NaCl}\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaOH}=0,3\cdot40=12\left(g\right)\\m_{NaCl}=0,3\cdot58,5=17,55\left(g\right)\end{matrix}\right.\)

Bảo toàn khối lượng: \(m_{kếttủa}=m_{muối}+m_{NaOH}-m_{NaCl}=7,16\left(g\right)\)