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True [87]
3 years ago
11

The first task you are assigned in your summer laboratory job is to prepare a concentrated NaOH stock solution. The molecular we

ight of NaOH is 40. How many grams of solid NaOH will you need to weigh out to obtain a 500 mL solution that has a concentration of 10 M
Chemistry
1 answer:
Jet001 [13]3 years ago
6 0

Answer:

200g of NaOH

Explanation:

The following data were obtained from the question:

Molar mass of NaOH = 40g/mol

Volume = 500mL

Molarity of NaOH = 10M

Next, we shall determine the number of mole of NaOH present in the solution. This is illustrated below:

Molarity of NaOH = 10M

Volume = 500mL = 500/1000 = 0.5L

Mole of NaOH =...?

Molarity = mole /Volume

10 = mole /0.5

Cross multiply

Mole = 10 x 0.5

Mole of NaOH = 5 moles

Finally, we can convert 5 moles of NaOH to grams. This is illustrated below:

Molar mass of NaOH = 40g/mol

Mole of NaOH = 5 moles

Mass of NaOH =...?

Mole = Mass /Molar Mass

5 = Mass /40

Cross multiply

Mass = 5 x 40

Mass of NaOH = 200g

Therefore, 200g of NaOH is needed to prepare the solution.

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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.

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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.

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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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