Answer:
6.43 moles of NF₃.
Explanation:
The balanced equation for the reaction is given below:
N₂ + 3F₂ —> 2NF₃
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Finally, we shall determine the number of mole of nitrogen trifluoride (NF₃) produced by the reaction of 9.65 moles of Fluorine gas (F₂). This can be obtained as follow:
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Therefore, 9.65 moles of F₂ will react to to produce = (9.65 × 2)/3 = 6.43 moles of NF₃.
Thus, 6.43 moles of NF₃ were obtained from the reaction.
The grams of sodium chloride that will be made is 292.5 g
<u><em>calculation</em></u>
2Na +Cl₂ → 2NaCl
step 1: calculate the moles of Na
moles = mass/molar mass
From periodic table the molar mass of Na = 23 g/mol
moles = 115 g/23 g /mol = 5 moles
Step 2 : use the mole ratio to determine the moles of NaCl
Na:NaCl is 2:2 = 1:1 therefore the moles of NaCl is also= 5 moles
Step 3: find the mass of NaCl
mass= moles x molar mass
The molar mass of NaCl = 23 + 35 .5 =58.5 g/mol
mass= 5 moles x 58.5 g/mol =292.5 g
1. To step
2. To throw
3. Measure
4. To join
5. Place
5.70 g ZrF4* (1 mol ZrF4/ 167.2 g ZrF4)= 0.0341 mol ZrF4.
The final answer is 0.0341 mol ZrF4~
