The mass of NaCl needed for the reaction is 91.61 g
We'll begin by calculating the number of mole of F₂ that reacted.
- Gas constant (R) = 0.0821 atm.L/Kmol
PV = nRT
1.5 × 12 = n × 0.0821 × 280
18 = n × 22.988
Divide both side by 22.988
n = 18 / 22.988
n = 0.783 mole
Next, we shall determine the mole of NaCl needed for the reaction.
F₂ + 2NaCl —> Cl₂ + 2NaF
From the balanced equation above,
1 mole of F₂ reacted with 2 moles of NaCl.
Therefore,
0.783 mole F₂ will react with = 0.783 × 2 = 1.566 moles of NaCl.
Finally, we shall determine the mass of 1.566 moles of NaCl.
- Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass = mole × molar mass
Mass of NaCl = 1.566 × 58.5
Mass of NaCl = 91.61 g
Therefore, the mass of NaCl needed for the reaction is 91.61 g
Learn more about stiochoimetry: brainly.com/question/25830314
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The given question is incomplete. The complete question is as follows.
A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate forms. The precipitate is filtered and dried and then found to have a mass of 212 mg. What mass of barium was in the original solution? (Assume that all of the barium was precipitated out of solution by the reaction.)
Explanation:
When
and
are added then white precipitate forms. And, reaction equation for this is as follows.
It is given that mass (m) is 212 mg or 0.212 g (as 1 g = 1000 mg). Molecular weight of
is 233.43.
Now, we will calculate the number of moles as follows.
No. of moles = mass × M.W
= 
= 0.00091 mol of
Hence, it means that 0.00091 mol of
. Now, we will calculate the mass as follows.
Mass = moles × MW
=
= 0.124 grams or 124 mg of barium
Thus, we can conclude that mass of barium into the original solution is 124 mg.