Đáp án:

 1. a. V = 16,8 lít

b. \(C\%_{CaCl_2}=18,5\%\)

2. \(m_{Cu}=3,2\ \text{gam}\)

Giải thích các bước giải:

Bài 1.

a. \(n_{Ca}=\dfrac{30}{40}=0,75\ \text{mol}\\ Ca+2H_2O\to Ca(OH)_2+H_2\\ Ca(OH)_2+MgCl_2\to CaCl_2+Mg(OH)_2\\ Mg(OH)_2+2HCl\to MgCl_2+2H_2O\)

\(\to n_{H_2}=n_{Ca}=0,75\ \text{mol}\to V_{H_2}=0,75\cdot 22,4=16,8\ \text{lít}\)

\(n_{HCl}=\dfrac{400\cdot 6,3875\%}{36,5}=0,7\ \text{mol}\to n_{Mg(OH)_2}=\dfrac{0,7}2=0,35\ \text{mol}\to n_{MgCl_2}=n_{Mg(OH)_2}=0,35\ \text{mol}\to m_{dd\ MgCl_2}=\dfrac{0,35\cdot 95}{19\%}=175\ \text{gam}\)

\(n_{Ca(OH)_2}=n_{Ca}=0,75\ \text{mol}\to m_{Ca(OH)_2}=55,5\ \text{gam}\)

\(\to m_A=55,5+175-0,35\cdot 58=210,2\ \text{gam}\)

\(n_{CaCl_2}=n_{Mg(OH)_2}=0,35\ \text{mol}\to C\%_{CaCl_2}=\dfrac{0,35\cdot 111}{210,2}\cdot 100\%=18,5\%\)

Bài 2. 

Khối lượng lá sắt tăng lên là: 

\[\Delta_m=10,4-10=0,4\ \text{gam}\]

PTHH: \(Fe+ CuSO_4\to FeSO_4+Cu\)

Theo PTHH: 1 mol Cu → 1 mol Fe \(⇒Δ'_m=64-56=8\ \text{gam}\)

mà Δm = 0,4 gam \(⇒ n_{Cu}=\dfrac{0,4\cdot 1}{8}=0,05\ \text{mol}\)

\(⇒m_{Cu}=64\cdot 0,05=3,2\ \text{gam}\)