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Svetradugi [14.3K]
2 years ago
5

A 150 mL sample of hydrochloric acid (HCl) completely reacted with 60.0 mL of a 0.100 M NaOH solution. The equation for the reac

tion is given below. HCl + NaOH Right arrow. NaCl + H2O What was the original concentration of the HCl solution? 0.040 M 0.25 M 1.50 M 2.50 M
Chemistry
2 answers:
iragen [17]2 years ago
8 0

Answer:

0.040M

Explanation:

just took the review

goblinko [34]2 years ago
5 0

Answer:

0.040M

Explanation:

The following data were obtained from the question:

Volume of acid (Va) = 150mL

Volume of base (Vb) = 60mL

Molarity of base (Mb) = 0.1M

Molarity of acid (Ma) =..?

Next, the balanced equation for the reaction. This is given below:

HCl + NaOH —> NaCl + H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 1

Finally, we shall determine the concentration of the acid, HCl as follow:

MaVa / MbVb = nA/nB

Ma x 150 / 0.1 x 60 = 1/1

Cross multiply to express in linear form

Ma x 150 = 0.1 x 60

Divide both side by 150

Ma = 0.1 x 60 / 150

Ma = 0.04M

Therefore, the concentration of the acid is 0.04M

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What is the limiting reagent when a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen?​
UNO [17]

Answer:

Ammonia is limiting reactant

Amount of oxygen left  = 0.035 mol

Explanation:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Balance chemical equation:

4NH₃ + 3O₂     →     2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

                      NH₃          :            N₂

                        4             :             2

                      0.12           :           2/4×0.12 = 0.06

                      NH₃         :            H₂O

                        4            :             6

                        0.12       :           6/4×0.12 = 0.18

                       

                       O₂            :            N₂

                        3             :             2

                      0.125        :           2/3×0.125 = 0.08

                        O₂           :            H₂O

                        3              :             6

                        0.125       :           6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

                        NH₃          :             O₂

                           4            :              3

                           0.12       :          3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left  = 0.125 - 0.09 = 0.035 mol

3 0
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