MNaHCO₃: 23+1+12+(48×3) = 84g
mCH₃COOH: 12+(1×3)+12+(16×2)+1 = 60g
.................
84g NaHCO₃ react with 60g CH₃COOH
83g NaHCO₃ react with...........
84g ----- 60g
83g ----- X
X = 59,29g CH₃COOH
We used 70g CH₃COOH, it' too much.
So, acetic acid is excess reagent, and sodium bicarbonate is limiting reagent.
_______________________________
B) Amount of CH3COOH is in excess.
:•)
The question is incomplete, the complete question is; Describe the preparation of 2.00 L of 0.108 M BaCl2 from BaCl2.2H2O
Answer:
See explanation
Explanation:
First we must know the molar mass of the compound = 244.3 g/mol
Now we must calculate the number of moles of BaCl2.2H2O in the solution.
number of moles = concentration * volume = 0.108 M * 2.00 L = 0.216 moles
Now the mass of solid that we need to take can be obtained from;
number of moles = mass/molar mass
mass = number of moles * molar mass
mass = 0.216 moles * 244.3 g/mol
Mass = 52.8 g
Hence, we must weigh 52.8 g of BaCl2.2H2O accurately in a balance and dissolve it in 2 L of water to give 0.108 M solution of BaCl2.2H2O.
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Answer:
The amount of precipitate formed would 7.175 grams of silver chloride.
Explanation:
Moles of NaCl = n
Volume of NaCl solution = 50.0 mL = 0.050 L
Molarity of the hydrogen peroxide = 2.0 M
Moles of silver nitarte = n'
Volume of silver nitrate solution = 50.0 mL = 0.050 L
Molarity of the silver nitrate = 1.0 M
According to reaction, 1 mole of of silver nitrate reacts with 1 mole of NaCl. Then 0.050 mole of silve nitrate will :
of NaCl
This means that silver nitrate is in limiting amount and NaCl is in excessive amount.
So, the amount of AgCl depends upon amount of silver nitrate.
According to reaction, 1 mole of silver nitrate gives 1 mole of AgCl.
Then 0.050 moles of silver nitrate will give;
of AgCl
Mass of 0.050 moles of AgCl ;
The amount of precipitate formed would 7.175 grams of silver chloride.
Molar mass KCl = <span>74.5513 g/mol
Number of moles:
21.9 / 74.5513 => 0.293 moles
Volume = 869 mL / 1000 => 0.869 L
Molarity = moles / Volume
Molarity = 0.293 / 0.869
=> 0.337 M</span>
