What will be the volume of chlorine gas at S.T.P. produced during electrolysis of water molten magnesium chloridewhich produces 6.5 gram of mg
Answers
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From the formula of magnesium chloride (MgCl2):
One mole of MgCl2 gives 1 mole of Mg (24g) and one mole of Chlorine gas (Cl2)
One mole of gas occupies 22.4 liters (22400 cm³) at S.T.P.
⇒ 24g of Mg ≡ 22400 cm³ of Chlorine gas at S.T.P.
∴ 6.5 g of Mg ≡ (22400 x 6.5) ÷ 24
≡ 6066.67 cm³
One mole of MgCl2 gives 1 mole of Mg (24g) and one mole of Chlorine gas (Cl2)
One mole of gas occupies 22.4 liters (22400 cm³) at S.T.P.
⇒ 24g of Mg ≡ 22400 cm³ of Chlorine gas at S.T.P.
∴ 6.5 g of Mg ≡ (22400 x 6.5) ÷ 24
≡ 6066.67 cm³
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Answer:
During electrolysis of water molten magnesium chloride (MgCl₂) which produces 6.5 grams of Mg, the volume of chlorine gas produced at S.T.P will be 5.6 L.
Explanation:
One mole of any gas at the S.T.P has a 22.4 L volume.
One mole of molten magnesium chloride (MgCl₂) produces 1 mole of Mg (24 gms) and 1 mole of Chlorine gas (22.4 L).
If 6 gms of Mg is produced, then how many liters of chlorine gas is produced.
Calculations -
24 gms of Mg = 22.4 L of chlorine gas at S.T.P
6 gms of Mg = (22.4/24) × 6 L = 5.6 L of chlorine gas at S.T.P
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