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
The answer is A lithium sulfite
The answer is
<span>2PbS(s) + 3O2(g) = 2PbO(s) + 2SO2(g)
Your answer is not yet balanced because you have 3 oxygen atoms. it should be balanced by multiplying both side by 2 such as the balanced equation I made. To check it, I will explain why your answer is not yet balanced.
check: (from your equation)
</span> 1-Pb-1
1-S-1
2 -O-3
the difference between the reactant and the product of Oxygen will prove that it is not yet balanced.
If you use 2PbS(s) + 3O2(g) = 2PbO(s) + 2SO2(g), to check it:
2-Pb-2
2-S-2
6 -O-6
then this is now balance
