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Mumz [18]
2 years ago
13

If 4500 g of Zr metal is reacted with 1.5 x 104 g of water what is the limiting reactant?

Chemistry
1 answer:
PIT_PIT [208]2 years ago
6 0

Answer:

Zr

Explanation:

The balance equation of the reaction is:

Zr + 2 H₂O ⇒ ZrO₂ + 2 H₂

Being:

  • Zr: 91 g/mole
  • H: 1 g/mole
  • O: 16 g/mole

the molar mass of the reactants participating in the reaction are:

  • Zr: 91 g/mole
  • H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole

If by stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction) they react in moles:

  • Zr: 1 mole
  • H₂O: 2 moles

then by stoichiometry the following quantity of mass react:

  • Zr: 1 mole*91 g/mole= 91 g
  • H₂O: 2 moles*18 g/mole=  36 g

Now the following rule of three applies: if 91 g of Zr react with 36 g of H₂O, 4500 g of Zr with how much mass of H₂O would it react?

mass of H_{2} O=\frac{4500 grams of Zr*36 grams of H_{2}O }{91 grams of Zr}

mass of H₂O= 1,780 grams

But 1780 grams of H₂0 are not available, 1.5 * 10⁴(15,000) grams are available. Since it has more mass than it needs to react with 4500 grams of Zr, Zr zirconia will be the limiting reagent.

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Explanation:

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(1) Microwave  =   1 m to 1 mm = 10^9 nm to 10^6 nm

(2) Ultraviolet  =    10 nm to 400 nm

(3) Radio waves  =   1 mm to 100 km = 10^6 nm to 10^{14}nm

(4) Infrared  =    700 nm to 1 mm

(5) X-ray  =   0.01 nm to 10 nm

(6) Visible =   400 nm t0 700 nm

a) In order of increasing wavelength:

: 5 < 2 < 6 < 4 < 1 < 3

b) Frequency of the electromagnetic wave given as:

\nu=\frac{c}{\lambda }

\nu = frequency

\lambda = Wavelength

c = speed of light

\nu \propto \frac{1}{\lambda }

So, the increasing order of frequency:

: 3 < 1 < 4 < 6 < 2 < 5

c) Energy(E) of the electromagnetic wave is given by Planck's equation :

E=h\nu

E\propto \nu

So, the increasing order of energy:

: 3 < 1 < 4 < 6 < 2 < 5

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Be sure to answer all parts. Nitric oxide, NO·, is a radical thought to cause ozone destruction by a mechanism similar to that o
SIZIF [17.4K]

Answer:

See image attached and explanation

Explanation:

The stratospheric ozone layer is very important in absorbing high-energy ultraviolet radiation that is harmful to living systems on earth. The concentration of ozone in the stratosphere is determined by both thermal and photochemical pathways for its decomposition. Nitric oxide, NO, is a trace constituent in the stratosphere that reacts with ozone to form nitrogen dioxide, NO2, and the diatomic oxygen molecule. The nitrogen-oxygen bond in NO2 is relatively weak. When an NO2 molecule encounters an oxygen atom, it transfers an oxygen, forming O2 and NO. The chemical reactions involved are formations of NO2 following by reaction of NO2 with atomic oxygen for form NO and O2. The sum of both reactions show that the overall reaction is simply the reaction of ozone with atomic oxygen to form two molecules of molecular oxygen. Hence, NO only serves as a catalyst, it does not undergo a permanent change itself.

6 0
3 years ago
A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products h
statuscvo [17]

Explanation:

First, calculate the moles of CO_{2} using ideal gas equation as follows.

                PV = nRT

or,          n = \frac{PV}{RT}

                = \frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}      (as 1 bar = 1 atm (approx))

                = 0.183 mol

As,   Density = \frac{mass}{volume}

Hence, mass of water will be as follows.

                Density = \frac{mass}{volume}

             0.998 g/ml = \frac{mass}{3.26 ml}    

                 mass = 3.25 g

Similarly, calculate the moles of water as follows.

        No. of moles = \frac{mass}{\text{molar mass}}

                              =  \frac{3.25 g}{18.02 g/mol}            

                              = 0.180 mol

Moles of hydrogen = 0.180 \times 2 = 0.36 mol

Now, mass of carbon will be as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

          0.183 mol =  \frac{mass}{12 g/mol}            

                              = 2.19 g

Therefore, mass of oxygen will be as follows.

              Mass of O = mass of sample - (mass of C + mass of H)

                                = 3.50 g - (2.19 g + 0.36 g)

                                = 0.95 g

Therefore, moles of oxygen will be as follows.

          No. of moles = \frac{mass}{\text{molar mass}}

                               =  \frac{0.95 g}{16 g/mol}            

                              = 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

                         C              H           O

No. of moles:  0.183        0.36       0.059

On dividing:      3.1           6.1            1

Therefore, empirical formula of the given compound is C_{3}H_{6}O.

Thus, we can conclude that empirical formula of the given compound is C_{3}H_{6}O.            

6 0
3 years ago
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