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defon
2 years ago
13

Concentrated hydrochloric acid is 36% (Weight/Volume) hydrochloric acid and has a density of 1.18 g.cm -3.

Chemistry
1 answer:
Juli2301 [7.4K]2 years ago
6 0

i. The new concentration is 0.116 M

ii. The new concentration is 0.029 M

<h3>What is the concentration of the original hydrochloric acid acid?</h3>

The concentration of the original or stock hydrochloric acid is calculated as follows:

  • Molarity = Percentage concentration * Density * 1000/Molar mass * 100

Molarity of stock HCl = 36 * 1.18* 1000 /36.5 * 100

Molarity of stock HCl = 11.6 mol/dm

i. Using the dilution formula: M₁V₁ = M₂V₂

M₂ = M₁V₁ /V₂

M₂ = 11.6 * 10/1000

M₂ = 0.116 M

The new concentration = 0.116 M

ii. Using M₁V₁ = M₂V₂

M₂ = M₁V₁ /V₂

M₂ = 0.116 * 5/20

M₂ = 0.029 M

The new concentration = 0.029 M

In conclusion, the new concentrations are found using the dilution formula.

Learn more about molarity at: brainly.com/question/17138838

#SPJ1

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Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

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