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pishuonlain [190]
2 years ago
9

Consider the following balanced equation. SiO2(s)+3C(s)→SiC(s)+2CO(g) Complete the following table, showing the appropriate numb

er of moles of reactants and products. If the number of moles of a reactant is provided, fill in the required amount of the other reactant, as well as the moles of each product formed. If the number of moles of a product is provided, fill in the required amount of each reactant to make that amount of product, as well as the amount of the other product that is made. molSiO2 molC molSiC molCO _____ 9 _____ _____ 1 _____ _____ _____ _____ _____ _____ 26 _____ 7.5 _____ _____ 1.4 _____ _____ _____ Part A Complete the first row.
Chemistry
1 answer:
vlada-n [284]2 years ago
5 0

Answer:

mol(SiO₂)              mol(C)               mol(SiC)                    mol(CO)

      3                          9                          3                                6

      1                           3                           1                                2

     13                         39                         13                             26

    2.5                        7.5                       2.5                            5.0

    1.4                         4.2                        1.4                            2.8

Explanation:

  • From the balanced equation:

<em>SiO₂(s) + 3C(s) → SiC(s) + 2CO(g),</em>

  • It is clear that 1.0 mole of SiO₂ reacts with 3.0 moles of C to produce 1.0 mole of SiC and 2.0 moles of CO.
  • We can complete the table of no. of moles of each component:

<u><em>1. 9.0 moles of C:</em></u>

We use the triple amount of C, so we multiply the others by 3.0.

So, it will be 3.0 moles of SiO₂ with 9.0 moles of C that produce 3.0 moles of SiC and 6.0 moles of CO.

<u><em>2. 1.0 mole of SiO₂:</em></u>

We use the same amount of SiO₂ as in the balnced equation, so the no. of moles of other components will be the same as in the balanced equation.

So, it will be 1.0 moles of SiO₂ with 3.0 moles of C that produce 1.0 moles of SiC and 2.0 moles of CO.

<u><em>3. 26.0 moles of CO:</em></u>

We use the amount of CO higher by 13 times than that in the balanced equation, so we multiply the others by 13.0.

So, it will be 13.0 moles of SiO₂ with 39.0 moles of C that produce 13.0 moles of SiC and 26.0 moles of CO.

<u><em>4. 7.5 moles of C:</em></u>

We use the amount of C higher by 2.5 times than that in the balanced equation, so we multiply the others by 2.5.

So, it will be 2.5 moles of SiO₂ with 7.5 moles of C that produce 2.5 moles of SiC and 5.0 moles of CO.

<u><em>5. 1.4 moles of SiO₂:</em></u>

We use the amount of SiO₂ higher by 1.4 times than that in the balanced equation, so we multiply the others by 1.4.

So, it will be 1.4 moles of SiO₂ with 4.2 moles of C that produce 1.4 moles of SiC and 2.8 moles of CO.

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Answer:

Hg(NO3)2 + Na2CO3 --> 2NaNO3 + HgCO3

4 0
2 years ago
When the temperature of a 3. 0-l sample of a gas is dropped from 200°c to 100°c, what will be the final volume of the gas sample
ipn [44]

P1V1T1=P2V2T2 Add 273 to convert degrees Celsius to Kelvin:

∴200×25/298=250×V2/273, ∴V2=200×25×273/298×250, ∴V2=18.32L

<h3>Where is the volume equation?</h3>

The basic formula for volume is length, breadth, and height, as opposed to length, width, and height for the area of a rectangular shape. The calculation is unaffected by how you refer to the various dimensions; for instance, you can use "depth" instead of "height."

<h3>What is chemistry using volume units?</h3>

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4 0
1 year ago
Consider the reaction of solid aluminum iodide and potassium metal to form solid potassium iodide and aluminum metal.The balance
ratelena [41]

Answer:

674.26 g of AlI₃

Explanation:

We'll begin by calculating the theoretical yield of aluminum (Al). This can be obtained as follow:

Percentage yield of Al = 67.8%

Actual yield of Al = 30.25 g

Theoretical yield of Al =?

Percentage yield = Actual yield /Theoretical yield × 100/

67.8% = 30.25 / Theoretical yield

67.8 / 100 = 30.25 / Theoretical yield

0.678 = 30.25 / Theoretical yield

Cross multiply

0.678 × Theoretical yield = 30.25

Divide both side by 0.678

Theoretical yield = 30.25 / 0.678

Theoretical yield of Al = 44.62 g

Next, we shall determine the mass of AlI₃ that reacted and the mass of Al produced from the balanced equation. This can be obtained as follow:

AlI₃(s) + 3K(s) → 3KI(s) + Al(s)

Molar mass of AlI₃ = 27 + (3×127)

= 27 + 381 = 408 g/mol

Mass of AlI₃ from the balanced equation = 1 × 408 = 408 g

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 1 × 27 = 27 g

Summary:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Finally, we shall determine the mass of

AlI₃ required to produce 44.62 g of Al. This can be obtained as follow:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Therefore, Xg of AlI₃ will react to produce 44.62 g of Al i.e

Xg of AlI₃ = (408 × 44.62)/27

Xg of AlI₃ = 674.26 g

Thus, 674.26 g of AlI₃ is needed for the reaction.

8 0
3 years ago
How much heat in j is giving out when 85.0g of lead cools from 200.0 c to 10.0c
photoshop1234 [79]

Answer:

q = - 2067.2 J of Heat is giving out when 85.0g of lead cools from 200.0 c to 10.0 c.

Explanation:

The Specific Heat capacity of Lead is 0.128 \frac{J}{g\ ^{0}C}

This means, increase in temperature of 1 gm of lead by 1 ^{0}\ C will require 0.128 J of heat.

Formula Used :

q = m.c.\Delta T

q = amount of heat added / removed

m = mass of substance in grams = 85.0 g

c = specific heat of the substance = 0.128

q = m.c.\Delta T = Change in temperature

                                          = final temperature - Initial temperature

                                          = 10 - 200

                                          = -190 ^{0}\ C

put value in formula

q = -  85\times 0.128\times 190

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q = - 2067.2 J

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