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 volume of a gas that is required yo react with 4.03 g mg at STP is 1856 ml
calculation/
- calculate the moles of Mg used
moles=mass/molar mass
moles of Mg is therefore=4.03 g/ 24.3 g/mol=0.1658 moles
- by use of mole ratio of Mg:O2 from the equation which is 2:1
the moles 02=0.1679 x1/20.0829 moles
- at STP 1 mole of a gas= 22.4 l
0.0895 moles=? L
- =0.0895 moles x22.4 l/ 1 mole=1.8570 L
into Ml = 1.8570 x1000=1856 ml approximately to 1860
None of the choices are right
In an oxidation-reduction reaction there is an exchange of electrons.
The exchange of electrons implies change in the oxidation states: at least one element increases its oxidation number while other reduces it.
By simple ispection you can predict that in the equation b. there is a change in oxidation states of Cl and Mn.
Now you can check it:
Equation 4H Cl + Mn O2 -> Mn Cl2 + 2H2 O + Cl2
oxidation sates 1+ 1- 4+ 2- 2+ 1- 1+ 2- 0
The oxidation state of Cl in HCl is 1- and it changed to 0 in Cl2
The oxidation state of Mn in MnO2 is 4+ and it changed to 2+ in MnCl2
Answer b.