calculate the amount of Agcl is formed by the action of 5.850 g of Nacl on an excess of AgNo3
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rabhiraj422gmailcom
17.06.2019
Chemistry
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what amount silver chloride is formed by action of 5.8 50 gram of sodium chloride all and excess of silver nitrate
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Explanation:
NaCl is limiting reagent
Amount of product , i.e AgCl depends on limiting reagent NaCl
1 mole of NaCl gives 1 mole of AgCl
molar mass of NaCl is 58.5
no of moles of NaCl available is 5.85/58.5 = 0.1
so 0.1 mole of AgCl will be formed
molar mass of AgCl is 143.5
so no. of moles * molarmass = mass
mass= 143.5*0.1
= 14.35g.
Hope this answer is helpful for us.
Answer:
Balanced chemical reaction is
NaCl + AgNO₃ ------→ AgCl + NaNO₃
Molar mass of NaCl = Atomic mass of Na + Atomic mass of Cl
= 23 + 35.5 = 58.50 g
Molar mass of AgNO₃ = Atomic mass of (Ag + N) + 3 (Atomic mass of O)
= 108 + 14 + 3(16) = 122 + 48 = 170 g
Molar mass of AgCl = Atomic mass of Ag + Atomic mass of Cl
= 108 + 35.5 = 143.5 g
From the above data, we can conclude that 58.50 g of sodium chloride (NaCl) react with 170 g of silver nitrate (AgNO₃).
5.850 g of NaCl react with x g of AgNO₃.
where x = (170/58.50)×5.850 = 17 g
Since the silver nitrate is present in excess. 170 g of AgNO₃ produces 143.5 g of AgCl.
1 g of AgNO₃ produces = 143.5/170 g of AgCl
∴ 17 g of AgNO₃ produces = (143.5/170)×17
= 14.35 g of AgCl
Hence, the amount of AgCl formed by the action of 5.850 g of NaCl on an excess of AgNO₃ is 14.35 g.
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