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True [87]
3 years ago
8

Magnesium ions can be precipitated from seawater by the addition of sodium hydroxide. How much sodium hydroxide in grams must be

added to a sample of seawater to completely precipitate the 88.4 mg of magnesium present?
Chemistry
1 answer:
Anna35 [415]3 years ago
8 0

Explanation:

Solution 86P

Here, we have to determine the amount of sodium hydroxide required to precipitate all of the Mg.

Mg2+(aq) +2 NaOH(aq) ------> Mg(OH)2 (s) + 2Na+(aq)

Therefore

Molar mass of NaOH =39.997 g/mol

Molar mass of Mg =24.305 g/mol

Thus,the conversion factors

g NaOH(aq)= mol NaOH(aq) = g Mg = mol Mg

Thus the amount of sodium hydroxide required

=  88.4 mg of  Mg x x x x...

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Predict the mass of iron (III) sulfide produced when 3.0 g of iron filings react completely with 2.5 g of yellow sulfur solid, S
777dan777 [17]
1) Chemical equation

16Fe(s) + 3S8(s) ---> 8Fe2S3

2) Molar ratios:

16 mol Fe : 3 mole S8 : 8 mol Fe2S3

3) Convert masses in grams to number of moles

number of moles = mass in grams / molar mass

a) iron, Fe

mass = 3.0 g
atomic mass = 55.845 g/mol

=> number of moles of Fe = 3.0g / 55.845 g/mol = 0.0537 mol

b) Sulfur, S8

mass = 2.5 g
molar mass = 8*32.065 g/mol = 256.52 g/mol

=> number of moles of S8 = 2.5g / 256.52 g/mol = 0.009746 mol

4) Limiting reactant

Theoretical ratio                           actual ratio

16 mol Fe / 3 mol S8                 0.0537 mol Fe / 0.009746 mol S8

5.33                                               5.50

So, there is a little bit more Fe than the theoretical needed to react all the S8, which means the S8 is the limiting reactant.

5) Calculate the number of moles of iron (III) produced with 2.5 g (0.009746 moles) of S8

3moles S8 / 8 moles Fe2S3 = 0.009746 moles S8 / x

=> x = 0.009746 * 8 / 3 moles Fe2S3 = 0.026 moles Fe2S3

6) Convert 0.026 moles Fe2S3 into grams

mass in grams = number of moles * molar mass

molar mass of Fe2S3 = 207.9 g/mol

mass = 0.026 mol * 207.9 g/mol = 5.40 g

7) Answer: option D)




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What mass of carbon dioxide gas occupies a volume of 81.3L at 204 kPa and a temperature of 95.0°C?
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Answer:

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Given parameters:

Volume of gas  = 81.3L

Pressure of gas = 204kPa

temperature of gas = 95°C

Unknown:

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Solution:

To solve this problem, the ideal gas law will be well suited. The ideal gas law is a fusion of Boyle's law, Charles's law and Avogadro's law.

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Answer:

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