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user100 [1]
3 years ago
5

How many moles in one gram of Calcium ??

Chemistry
1 answer:
kirill [66]3 years ago
8 0

Answer:

0.0249 moles in 1 g of Ca

Explanation:

Let's think in the molar mass of Ca.

Ca = 40.08 g/mol

So 1 mol weighs 40.08 grams, or in the opposite 40.08 grams is the weigh of 1 mol

The rule of three will be:

40.08 g are contained in 1 mol

1 g may be contained in (1 . 1) / 40.08 = 0.0249 moles

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What is the mole fraction, x, of solute and the molality, m (or b), for an aqueous solution that is 19.0% naoh by mass?
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We will assume that the solvent is water. So, if we have 100 grams of the solution, 19 grams will be sodium hydroxide, while the remaining 81 grams will be water.

The molar weight of sodium hydroxide, NaOH, is 40. The molar weight of water is 18. Finding the moles of each:

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The mole fraction of NaOH is 0.0955
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15.00 grams of Chromium react with 15.00 grams of hydrobromic acid. Calculate the theoretical yield of the reaction. At STP what
s2008m [1.1K]

Answer:

(a) 18.03 g

(b) 2.105 L

(c) 85.15 %

Step-by-step explanation:

We have the masses of two reactants, so this is a<em> limiting reactant problem.  </em>

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

<em>Step 1</em>. <em>Gather all the information</em> in one place with molar masses above the formulas and masses below them.  

M_r:        52.00   80.91       291.71

                2Cr  +  6HBr ⟶ 2CrBr₃ + 3H₂

Mass/g:  15.00    15.00  

<em>Step 2</em>. Calculate the <em>moles of each reactant</em>  

  Moles of Cr = 15.00 × 1/52.00

  Moles of Cr = 0.2885 mol Cr

Moles of HBr = 15.00 × 1/80.91

Moles of HBr = 0.1854 mol HBr ×  

<em>Step 3</em>. Identify the<em> limiting reactant</em>  

Calculate the moles of CrCl₃ we can obtain from each reactant.  

<em>From Cr</em>:

The molar ratio of CrBr₃:Cr is 2 mol CrBr₃:2 mol Cr

Moles of CrBr₃ = 0.2885 × 2/2

Moles of CrBr₃ = 0.2885 mol CrCl₃

<em>From HBr: </em>

The molar ratio of CrBr₃:HBr is 2 mol CrBr₃:6 mol HBr.

Moles of CrBr₃ = 0.1854 × 2/6

Moles of CrBr₃ = 0.061 80 mol CrBr₃

The limiting reactant is HBr because it gives the smaller amount of CrBr₃.

<em>Step 4</em>. Calculate the <em>theoretical yields</em> of CrBr₃ and H₂.

Theoretical yield of CrBr₃ = 0.061 80 × 291.71/1

Theoretical yield of CrBr₃ = 18.03 g CrCl₃

The molar ratio is 3 mol H₂:6 mol HBr

   Theoretical yield of H₂ = 0.1854 × 3/6

   Theoretical yield of H₂ = 0.092 70 mol H₂

<em>Step 5</em>. Calculate the <em>volume of H₂</em> at STP

STP is 1 bar and 0 °C.

The molar volume of a gas at STP is 22.71 L.

Volume = 0.092 70 × 22.71/1

Volume = 2.105 L

<em>Step 6</em>. Calculate the <em>percent yield </em>

       % Yield = actual yield/theoretical yield × 100 %

Actual yield = 15.35 g

       % yield = 15.35/18.03 × 100

       % yield = <em>85.15 % </em>

8 0
3 years ago
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