⇒ What mass of NaHCo₃ would contain 8 g of oxygen ?
⇒ Find the mass of ferric sulphate that contains N₀ atoms ?
Answers
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\textsf{⇒ Find the mass of ferric sulphate that contains N₀ atoms ?}⇒ Find the mass of ferric sulphate that contains N₀ atoms ?
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\textsf{↣ Molar weight of Fe=56 g}↣ Molar weight of Fe=56 g
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\textsf{↣ Molar weight of S=32 g}↣ Molar weight of S=32 g
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\textsf{↣ Molar weight of O2=32 g i.e.Molar weight of O=16 g}↣ Molar weight of O2=32 g i.e.Molar weight of O=16 g
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\textsf{↣ Therefore,molar weight of ferric sulphate is - > (2*56)+3*(32+4*16)=400 g}↣ Therefore,molar weight of ferric sulphate is - > (2*56)+3*(32+4*16)=400 g
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\textsf{↣ Now, 400 g of ferric sulphate is 1 mol}↣ Now, 400 g of ferric sulphate is 1 mol
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\textsf{↣ 1 g of ferric sulphate is 1/400 mol}↣ 1 g of ferric sulphate is 1/400 mol
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\textsf{↣40 g of ferric sulphate is (1/400)*40=0.1 mol}↣40 g of ferric sulphate is (1/400)*40=0.1 mol
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\textsf{↣ So,40 g of ferric sulphate has 0.1 mol of it.}↣ So,40 g of ferric sulphate has 0.1 mol of it.
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Q)What mass of NaHCo₃ would contain 8 g of oxygen ?
The mass of residue, is CH3
The mass of residue, is CH3
The mass of residue, is CH3 COONa solution, should be 24g of CH3
The mass of residue, is CH3 COONa solution, should be 24g of CH3
The mass of residue, is CH3 COONa solution, should be 24g of CH3 COONa solution.
The mass of residue, is CH3 COONa solution, should be 24g of CH3 COONa solution.In this reaction, sodium bicarbonate reacts with acetic acid to produce sodium acetate, carbon dioxide and water
The mass of residue, is CH3 COONa solution, should be 24g of CH3 COONa solution.In this reaction, sodium bicarbonate reacts with acetic acid to produce sodium acetate, carbon dioxide and waterNaHCO3+CH 3COOH⟶CH3COONa+CO2+H2O
2+H2OAccording to law of conversation of masses,
2+H2OAccording to law of conversation of masses,total masses of reactants = total masses of products reactants are 8.4g NaHCO3and 20g CH3COOH.
and 20g CH3COOH.total mass of reactants is =28.4g
and 20g CH3COOH.total mass of reactants is =28.4gmass of residue = (28.4g reactants mass) − (4.4g CO )2 = 24.0g residue (sodium acetate solution)
2 = 24.0g residue (sodium acetate solution)mass of residue is thus 24 grams.
Q)Find the mass of ferric sulphate that contains N₀ atoms ?
Molar weight of Fe=56 g
Molar weight of Fe=56 gMolar weight of S=32 g
Molar weight of Fe=56 gMolar weight of S=32 gMolar weight of O2=32 g i.e.Molar weight of O=16 g
Molar weight of Fe=56 gMolar weight of S=32 gMolar weight of O2=32 g i.e.Molar weight of O=16 gTherefore,molar weight of ferric sulphate is -> (2*56)+3*(32+4*16)=400 g
Molar weight of Fe=56 gMolar weight of S=32 gMolar weight of O2=32 g i.e.Molar weight of O=16 gTherefore,molar weight of ferric sulphate is -> (2*56)+3*(32+4*16)=400 gNow, 400 g of ferric sulphate is 1 mol
Molar weight of Fe=56 gMolar weight of S=32 gMolar weight of O2=32 g i.e.Molar weight of O=16 gTherefore,molar weight of ferric sulphate is -> (2*56)+3*(32+4*16)=400 gNow, 400 g of ferric sulphate is 1 mol1 g of ferric sulphate is 1/400 mol
Molar weight of Fe=56 gMolar weight of S=32 gMolar weight of O2=32 g i.e.Molar weight of O=16 gTherefore,molar weight of ferric sulphate is -> (2*56)+3*(32+4*16)=400 gNow, 400 g of ferric sulphate is 1 mol1 g of ferric sulphate is 1/400 mol40 g of ferric sulphate is (1/400)*40=0.1 mol
Molar weight of Fe=56 gMolar weight of S=32 gMolar weight of O2=32 g i.e.Molar weight of O=16 gTherefore,molar weight of ferric sulphate is -> (2*56)+3*(32+4*16)=400 gNow, 400 g of ferric sulphate is 1 mol1 g of ferric sulphate is 1/400 mol40 g of ferric sulphate is (1/400)*40=0.1 molSo,40 g of ferric sulphate has 0.1 mol of it.