Question

1,cho 5.6g Fe tác dụng dd H₂SO₄ 9,8% . tính C%

2, cho 3,6g Mg vào 500g dd HCl 3,65% .tính C%

Answer 1

Bài 1: Fe + H₂SO₄ -> FeSO₄ + H₂ (1)

n_Fe = \(\dfrac{5,6}{56}\) = 0,1 (mol)

Theo PT (1) ta có:

+) n_{\(FeSO_4\)} = n_Fe = 0,1(mol) => m_{\(FeSO_4\)} = 0,1.152 = 15,2(g)

+) n_{\(H_2SO_4\)} = n_Fe = 0,1(mol) => m_{\(H_2SO_4\)} = 0,1.98 = 9,8(g)

=> m_dd H₂SO₄ 9,8% là: \(\dfrac{9,8.100\%}{9,8\%}\) = 100(g)

+) n_{\(H_2\)} = n_Fe = 0,1(mol) => m_{\(H_2\)} = 0,1.2 = 0,2(g)

m_dd = 5,6 + 100 - 0,2 = 105,4(g) => C% = \(\dfrac{15,2}{105,4}.100\%\) \(\approx\) 14,42 %

Answer 2

Bài 2: Mg + 2HCl -> MgCl₂ + H₂ (1)

n_Mg = \(\dfrac{3,6}{24}\) = 0,15 (mol)

m_HCl = \(\dfrac{500.3,65}{100}\) = 18,25(g) => n_HCl = \(\dfrac{18,25}{36,5}\) = 0,5(mol)

So sánh tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,5}{2}\) => HCl dư, bài toán tính theo Mg

Theo PT (1) ta có:

+) n_{\(MgCl_2\)} = n_Mg = 0,15(mol) => m_{\(MgCl_2\)} = 0,15.95 = 14,25(g)

+) n_{\(H_2\)} = n_Mg = 0,15(mol) => m_{\(H_2\)} = 0,15.2 = 0,3(g)

m_dd = 3,6 + 500 - 0,3 = 503,3(g) => C% = \(\dfrac{14,25}{503,3}.100\%\) \(\approx\) 2,83%