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3 years ago

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Cl2 + 2na → 2nacl 1.5 moles of cl2 and 4.0 moles of na are combined and allowed to react. the limiting reactant will be _(i)

_, and the theoretical yield of nacl will be ___(ii)____.

1 answer:

cricket20 [7]3 years ago

7 0

Hope this helps you;):):

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$\boxed{\text{(a) 14 L; (b) 711 kg}}$

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$\text{Volume} =8.5 \times 10^{7} \text{gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} = 3.22 \times 10^{8}\text{ L}\\\\\text{Milligrams of F}^{-} = 3.22 \times 10^{8}\text{ L} \times \dfrac{\text{ 1 mg F}^{-}}{\text{1 L}} = 3.22 \times 10^{8}\text{ mg F}^{-}\\\\\text{kilograms of F}^{-} = 3.22 \times 10^{8}\text{ mg F}^{-} \times \dfrac{\text{1 mg F}^{-}}{10^{6}\text{kg F⁻}} = \text{322 kg F}^{-}$

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Lewis structures for the perchlorate ion (ClO4−) can be drawn with all single bonds or with one, two, or three double bonds. Dra

Vanyuwa [196]

Answer:

The most important resonance structure is 4 (attached picture). Its bon order is $\frac{7}{4}$ or $1\frac{3}{4}$ .

Explanation:

A picture with 4 forms of the perchlorate structure is attached. The first structure has simple bonds. The second structure contains a double bond, the third structure has two double bonds and the fourth structure has three double bonds.

Formal charge = group number of the periodic table - number of bonds (number of bonding electrons / 2) - number of non-shared electrons (lone pairs)

The formal charges in the first structure is +3 in chlorine and -1 in oxygen.

The formal charges in the second structure is +2 in chlorine, -1 in oxygen and 0 in the double bond oxygen.

The formal charges in the third structure is +1 in chlorine, -1 in the single bond oxygens and 0 in the double bond oxygens.

The formal charges in the fourth structure is 0 in chlorine, -1 in the single bond oxygen and 0 in the double bond oxygens.

The most important resonance structure is given by:

Most atoms have 0 formal charge.

Lowest magnitude of formal charges.

If there is a negative formal charge, it's on the most electronegative atom.

Hence, the fourth structure is the mosr important.

The bond order of the structure is:

Total number of bonds: 7

Total number of bond groups: 4

Bond order= $\frac{7}{4} =1\frac{3}{4}$

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