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ASHA 777 [7]
3 years ago
14

Acid indigestion is sometimes neutralized with an antacid such as magnesium hydroxide (Mg(OH)2). What products will be released

when the antacid is mixed with the hydrochloric acid in the stomach?
a) H2O and OH
b) MgOH and H3O2
c) Cl(OH)2 and HNa
d) MgCl2 and H2O​
Chemistry
1 answer:
Lunna [17]3 years ago
3 0

Answer: d) MgCl_2 and H_2O​

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

A neutralization reaction is a double displacement reaction in which an acid reacts with base to produce salt and water. The H^+ ions from water reacts with OH^- ions from base to give water.

2HCl+Mg(OH)_2\rightarrow MgCl_2+2H_2O

Thus products that will be released when the antacid is mixed with the hydrochloric acid in the stomach are MgCl_2 and H_2O​

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Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of water is pro
Lubov Fominskaja [6]

This is an incomplete question, here is a complete question.

Aqueous sulfuric acid (H₂SO₄) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium sulfate (Na₂SO₄) and liquid water (H₂O) . If 12.5 g of water is produced from the reaction of 72.6 g of sulfuric acid and 77.0 g of sodium hydroxide, calculate the percent yield of water. Round your answer in significant figures.

Answer: The percent yield of water is, 46.8 %

Explanation : Given,

Mass of H_2SO_4 = 72.6 g

Mass of NaOH = 77.0 g

Molar mass of H_2SO_4 = 98 g/mol

Molar mass of NaOH = 40 g/mol

First we have to calculate the moles of H_2SO_4 and NaOH.

\text{Moles of }H_2SO_4=\frac{\text{Given mass }H_2SO_4}{\text{Molar mass }H_2SO_4}

\text{Moles of }H_2SO_4=\frac{72.6g}{98g/mol}=0.741mol

and,

\text{Moles of }NaOH=\frac{\text{Given mass }NaOH}{\text{Molar mass }NaOH}

\text{Moles of }NaOH=\frac{77.0g}{40g/mol}=1.925mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O

From the balanced reaction we conclude that

As, 1 mole of H_2SO_4 react with 2 mole of NaOH

So, 0.741 moles of H_2SO_4 react with 0.741\times 2=1.482 moles of NaOH

From this we conclude that, NaOH is an excess reagent because the given moles are greater than the required moles and H_2SO_4 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of H_2O

From the reaction, we conclude that

As, 1 mole of H_2SO_4 react to give 2 mole of H_2O

So, 0.741 moles of H_2SO_4 react to give 0.741\times 2=1.482 moles of H_2O

Now we have to calculate the mass of H_2O

\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O

Molar mass of H_2O = 18 g/mole

\text{ Mass of }H_2O=(1.482moles)\times (18g/mole)=26.68g

Now we have to calculate the percent yield of water.

\text{Percent yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Now put all the given values in this formula, we get:

\text{Percent yield}=\frac{12.5g}{26.68g}\times 100=46.8\%

Thus, the percent yield of water is, 46.8 %

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3 years ago
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