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1 year ago

What is observed when magnesium is reacted with hydrochloric acid? choose all that apply

Chemistry

1 answer:

Zanzabum1 year ago

3 0

It is observed that when magnesium reacts with hydrochloric acid, it produces visible bubbles of hydrogen gas.

A vigorous reaction will occur giving rise of heat as it is an exothermic reaction. If heat is applied then it should be more vigorous.

Reaction is as follows:

Mg(s) + HCl (aq) --> MgCl2 (aq) + H2(g)

Magnesium reacts easily with HCl to produce H2 gas and magnesium ions, Mg2+, and heat. The reaction is exothermic, so it heats up quickly.

Mg(s) + HCl (aq) --> MgCl2(aq) + H2(g)

The net ionic equation :

Mg(s) + 2H+ --> Mg2 + H2(g)

If water is removed from the solution then white crystals of Mgcl2 is obtained.

Or in simple words,

2Mg + 2HCl -> 2Mg+ + 2Cl- + H2(gas)

The magnesium is attacked by the hydrochloric acid resulting in the magnesium dissolving into the solution resulting in a solution of magnesium chloride in hydrochloric acid and the production of hydrogen gas.

Learn more about magnesium here : brainly.com/question/5759562

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Answer:

(1) The absorbance of the aspirin in solutions will increase.

(2) [ASA]f = 3.79x10⁻⁴M

(3) [ASA]i = 3.79x10⁻³M

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The Beer's Law is expressed by:

$A = \epsilon \cdot l \cdot C$ (1)

where A: is the absorbance of the species, ε: is the molar attenuation coefficient, l: is the pathlength and C: is the concentration of the species

(1) From equation (1), the relation between the absorbance of the species and its concentration is directly proportional, so if the aspirin concentration in solutions increases, the absorbance of the solutions will also increase.

(2) Starting in the given expression for the relationship between absorbance and concentration of ASA, we can calculate its concentration in the solution:

$A = 1061.5 \cdot [ASA]$

$[ASA] = \frac{A}{1061.5} = 3.79 \cdot 10^{-4}M$

Therefore, the aspirin concentration in the solution is 3.79x10⁻⁴ M

(3) To calculate the stock solution concentration, we can use the next equation:

$V_{i} [ASA]_{i} = V_{f} [ASA]_{f}$

where Vi: is the stock solution volume=10mL, Vf: is the solution diluted volume=100mL, [ASA]i: is the aspirin concentration of the stock solution and [ASA]f: is the aspirin concentration of the diluted solution

$[ASA]_{i} = \frac{V_{f} \cdot [ASA]_{f}}{V_{i}} = \frac {100mL \cdot 3.79\cdot 10^{-4} M}{10mL} = 3.79 \cdot 10^{-3} M$

Hence, the concentration of the stock solution is 3.79x10⁻³M

(4) To determine the aspirin mass in the tablet, we need to use the following equation:

$m_{ASA} = \eta_{ASA} \cdot M_{ASA} = [ASA]_{i} \cdot V_{0} \cdot M_{ASA}$

where η: is the aspirin moles = [ASA]i V₀, M: is the molar mass of aspirin=180.158g/mol, V₀: is the volume of the volumetric flask=250mL and [ASA]i: is the aspirin concentration in the volumetric flask which is equal to the stock solution=3.79x10⁻³M

$m_{ASA} = 3.79 \cdot 10^{-3} \frac{mol}{L} \cdot 0.250L \cdot 180.158 \frac{g}{mol} = 0.171 g$

Then, the aspirin mass in the tablet is 0.171 g.

I hope it helps you!

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