5.85g of NaCl was treated with concentrated H2SO4 and the gas evolved was passed into a solution of silver nitrate. The white precipitate obtained was filtered, dried and weighed. Assuming complete reaction, how many grams of precipitate was obtained?
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Here is the equation for the reaction between NaCl and H2SO4:
2 NaCl + H₂SO₄ ⇒2 HCl + Na₂SO₄
Calculate the moles of the NaCl used:
Moles = mass/ molar mass
mass = 5.85
molar mass = 58.5 g/mol
Moles = 5.85/58.5 = 0.1 moles
Use stochiometry to calculate the moles of the HCl gas produced by making use of the equation above.
From the above equation, the mole ratio between NaCl : HCl = 2:2 = 1:1
That means that the moles of the HCl produced are also 0.1 moles
The reaction between HCl and silver nitrate is as follows:
AgNO₃+ HCl ⇒ AgCl + HNO₃
The mole ratio between the HCl and the AgCl produced is 1:1, that means 0.1 moles of HCl will also produced 0.1 moles of AgCl.
Calculate the mass of 0.1 moles of AgCl:
mass = moles × molar mass
moles = 0.1moles
molar mass = 144.5g/mol
mass = 144.5 × 0.1 = 14.45g
Therefore the mass of AgCl produced ( the white precipitate) is 14.45g
2 NaCl + H₂SO₄ ⇒2 HCl + Na₂SO₄
Calculate the moles of the NaCl used:
Moles = mass/ molar mass
mass = 5.85
molar mass = 58.5 g/mol
Moles = 5.85/58.5 = 0.1 moles
Use stochiometry to calculate the moles of the HCl gas produced by making use of the equation above.
From the above equation, the mole ratio between NaCl : HCl = 2:2 = 1:1
That means that the moles of the HCl produced are also 0.1 moles
The reaction between HCl and silver nitrate is as follows:
AgNO₃+ HCl ⇒ AgCl + HNO₃
The mole ratio between the HCl and the AgCl produced is 1:1, that means 0.1 moles of HCl will also produced 0.1 moles of AgCl.
Calculate the mass of 0.1 moles of AgCl:
mass = moles × molar mass
moles = 0.1moles
molar mass = 144.5g/mol
mass = 144.5 × 0.1 = 14.45g
Therefore the mass of AgCl produced ( the white precipitate) is 14.45g
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