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AleksandrR [38]
3 years ago
13

How many grams of sodium chloride are present in a 0.75 M solution with a volume of 500.0 milliliters?

Chemistry
2 answers:
alexira [117]3 years ago
8 0
M = mol/l

So, M = 0.75 M
       V = 500.0 ML
        Mols = X
c = n/v
n = c(v)
n = 0.75(0.5)
   = 0.375mol
sodium Chloride is 58.5g/mol
1mol = 58.5g, 0.375mol(g)
0.375mol(58.5g)/1mol
Your answer would be: 21.94g




Hope that helps!!!



Yuri [45]3 years ago
6 0
My work and explanations are in the picture. I will comment with the rest of the work.

Hope this helps! :)

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a) The theoretical yield is 408.45g of BaSO_{4}

b) Percent yield = \frac{realyield}{408.45g}*100

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M=\frac{1.75molesBaCl_{2}}{1Lsolution}

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M=\frac{2.0moles[tex]Na_{2}SO_{4}}{1Lsolution}[/tex]

Therefore there are 2.0 moles of Na_{2}SO_{4}

2. Write the balanced chemical equation for the synthesis of the barium white pigment, BaSO_{4}:

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3. Determine the limiting reagent.

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As the BaCl_{2} is the smalles quantity, this is the limiting reagent.

4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.

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5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:

Percent yield = \frac{realyield}{theoreticalyield}*100

Percent yield = \frac{realyield}{408.45g}*100

The real yield is the quantity of barium white pigment you obtained in the laboratory.

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