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
Answer:
Please find the answer to the question below
Explanation:
In chemistry, the following mathematical formula is used to calculate the number of moles contained by a substance:
mole = mass of substance (g)/molar mass of substance (g/mol)
Molar mass of salicylic acid (C7H6O3) = 12(7) + 1(6) + 16(3)
= 84 + 6 + 48
= 138g/mol
Mass = 5.50grams
mole = 5.5/138
mole = 0.039
Approximately, the number of moles of 5.5grams of salicylic acid is 0.04moles. This is in accordance with the mole value (0.04) given in this question.
Explanation:
These elements are rare because:
<u>Helium fuses into the carbon by the combination of three helium nuclei (Z = 2) and one carbon nucleus (Z = 6), therefore bypassing elements with Z= 3, 4 and 5 which are lithium, beryllium, and boron respectively. Therefore, the fusion processes in cores of the stars do not form these three elements. </u>
Answer:
The Structure of "B" is alkene.
Explanation:
The compound "A" having R- configuration and undergoes Hofmann elimination to form an alkene.
The compound "B" on oxidatively cleaving with ozone followed by dimethyl sulfide forms 
and 
The structure of "A" and"B" is as follows.
The chemical reaction is expressed as:
2H2 + O2 = 2H2O
To determine the amount of oxygen used in the reaction, we use the amount of water produced and the relation of the substances in the reaction we do as follows:
209 g H2O ( 1 mol / 18.02 g ) ( 1 mol O2 / 2 mol H2O ) ( 32 g / 1 mol ) = 185.57 g O2
