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zubka84 [21]
2 years ago
5

In many species, a transition metal has an unusually high or low oxidation state. Write balanced equations for the following and

find the oxidation state of the transition metal in the product:(d) Vanadium(III) chloride reacts with Na metal under a COatmosphere to produce Na[V(CO)₆] and NaCl.
Chemistry
1 answer:
Liono4ka [1.6K]2 years ago
4 0

Transition metal, any of various chemical elements that have valence electrons—i.e., electrons that can participate in the formation of chemical bonds—in two shells instead of only one.

<h3>What are properties of transition metals?</h3>

The transition metals have the following physical properties in common:

  •  they are good conductors of heat and electricity.
  •    they can be hammered or bent into shape easily.
  •    they have high melting points (but mercury is a liquid at room temperature)
  •    they are usually hard and tough.
  •    they have high densities.

<h3>Where are the transition metals?</h3>

Early transition metals are on the left side of the periodic table from group 3 to group 7. Late transition metals are on the right side of the d-block, from group 8 to 11 (and 12 if it is counted as transition metals).

Learn more about transition metals here:

<h3>brainly.com/question/2426896</h3><h3 /><h3>#SPJ4</h3>
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Answer:

\boxed{\text{47.4 g}}

Explanation:

We are given the mass of two reactants, so this is a limiting reactant problem.

We know that we will need mases, moles, and molar masses, so, let's assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r:    17.03   32.00                 18.02  

           4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

m/g:     70.1      70.1

Step 1. Calculate the moles of each reactant

\text{Moles of CO } = \text{70.1 g} \times \dfrac{\text{1 mol}}{\text{17.03 g}} = \text{4.116 mol}\\\\\text{Moles of H$_{2}$O} = \text{70.1 g} \times \dfrac{\text{1 mol}}{\text{32.00 g}} = \text{2.191 mol}

Step 2. Identify the limiting reactant  

Calculate the moles of H₂O we can obtain from each reactant.

From NH₃:

The molar ratio of H₂O:NH₃ is 6:4.

\text{Moles of H$_{2}$O} = \text{4.116 mol NH$_{3}$} \times \dfrac{\text{6 mol H$_{2}$O}}{\text{4 mol NH$_{3}$}} = \text{6.174 mol H$_{2}$O}

From O₂:  

The molar ratio of H₂O:O₂ is 6:5.  

\text{Moles of H$_{2}$O} = \text{2.191 mol O$_{2}$} \times \dfrac{\text{6 mol H$_{2}$O}}{\text{5 mol O$_{2}$}} = \text{2.629 mol H$_{2}$O}

O₂ is the limiting reactant because it gives the smaller amount of H₂O.  

Step 3. Calculate the theoretical yield.

\text{Theor. yield } = \text{2.629 mol H$_{2}$O}\times \dfrac{\text{18.02 g H$_{2}$O}}{\text{1 mol H$_{2}$O}} = \textbf{47.4 g H$_{2}$O}\\\\\text{The maximum yield of H$_{2}$O is }\boxed{\textbf{47.4 g}}

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