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Sloan [31]
3 years ago
10

How many milliliters of a 5.0 M H2SO4 stock solution would you need to prepare 108.0 mL of 0.45 M H2SO4?

Chemistry
1 answer:
PIT_PIT [208]3 years ago
7 0
For the purpose we will use solution dilution equation:
c1xV1=c2xV2
Where, c1 - concentration of stock solution; V1 - a volume of stock solution needed to make the new solution; c2 - final concentration of new solution; V2 - final volume of new solution.
c1 = 5.00 M
c2 = 0.45 M
V1 = ?
V2 = 108 L
When we plug values into the equation, we get following:
5 x V1 = 0.45 x 108
<span>V1 = </span>9.72 L
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4 0
3 years ago
When 40.5 g of Al and 212.7 g of Cl2 combine in the reaction: 2 Al (s) + 3 Cl2 (g) --&gt; 2 AlCl3 (s) c) How many moles of the e
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Answer:

The answer to your question is 1.49 mol of Cl₂  

Explanation:

Data

mass of Al = 40.5 g

mass of Cl₂ = 212.7 g

moles of excess reactant = ?

- Balanced chemical reaction

                2 Al(s)  +  3Cl₂(g)  ⇒   2AlCl₃

Process

a) Calculate the molar mass of the reactants  

molar mass of Al = 2 x 26.98 = 53.96 g

molar mass of Cl₂ = 6 x 35.45 = 212.7 g

b) Calculate the theoretical proportion  Al/Cl₂ = 53.96/212.7 = 0.254

   Calculate the experimental proportion Al/Cl₂ = 40.5/212.7 = 0.19

As the experimental proportion is lower than the theoretical proportion we conclude that the limiting reactant is Aluminum.

c) Calculate the grams of excess reactant

                    53.96 g of Al ------------------ 212.7 g of Cl₂

                     40.5 g of Al -------------------  x

                      x = (40.5 x 212.7) / 53.96

                      x = 8614.35 / 53.96

                      x = 159.64 g of Cl₂

Excess Cl₂ = 212.7 - 159.64

                  = 53.057 g

d) Calculate the moles of Cl

                       35.45 g of Cl ----------------- 1 mol

                       53.057 g of Cl ---------------  x

                        x = (53.057 x 1)/35.45

                       x = 1.49 mol of Cl₂                    

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