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Ludmilka [50]
3 years ago
15

What is the concentration of an unknown Mg(OH)2 solution if it took an average of 15.4mL of

Chemistry
1 answer:
vova2212 [387]3 years ago
7 0

Answer:

0.077M

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2HCl + Mg(OH)2 —> MgCl2 + 2H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 2

The mole ratio of the base (nB) = 1

Step 2:

Data obtained from the question.

Concentration of base Cb =...?

Volume of base (Vb) = 10mL

Concentration of acid (Ca) = 0.1M

Volume of acid (Va) = 15.4mL

Step 3:

Determination of the concentration of the base, Mg(OH)2.

The concentration of the base can be obtained as follow:

CaVa/CbVb = nA/nB

0.1 x 15.4 /Cb x 10 = 2/1

Cross multiply to express in linear form

Cb x 10 x 2 = 0.1 x 15.4

Divide both side by 10 x 2

Cb = (0.1 x 15.4) /(10 x 2)

Cb = 0.077M

Therefore, the concentration of the base, Mg(OH)2 is 0.077M

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3 years ago
If 72.5 grams of calcium metal (Ca) react with 65.0 grams of oxygen gas (O2) in a synthesis reaction, how many grams of the exce
Natali5045456 [20]
2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)

65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.

Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
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=0.0282669621 g of O2 left over
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