Chemistry, 10.07.2019 03:30 hallmansean04

For each of the acid–base reactions, calculate the mass (in grams) of each acid necessary to completely react with and neutralize 4.85 g of the base. a. hcl(aq) + naoh(aq) s h2o(l) + nacl(aq) b. 2 hno3(aq) + ca(oh)2(aq) s 2 h2o(l) + ca(no3)2(aq) c. h2so4(aq) + 2 koh(aq) s 2 h2o(l) + k2so4(aq)

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Neutralization reaction is defined as the reaction of acid and base to yield salt and water. The strong acids and bases nullify each other's effect or neutralize each other.

(a) In the given reactions, the neutralization occurs as follows:

$\text {HCl}_{(aq)}\;+\; \text {NaOH}_{(aq)}\;\rightarrow\; \text{H}_{2}{\text O}_{(l)}\;+\;\text{NaCl}_{(aq)}$

In the above equation, 1 mol of NaOH completely reacts with 1 mol of HCl. The mass of NaOH is 4.85 grams, converting it into number of moles as:

$\text {n}&=\frac{\text m}{\text M}$

The molar mass of NaOH is = 39.997 g/mol, therefore,

$\text {n}&=\frac{\text 4.85} {39.997\text g/mol}$

=0.1201 mol.

Thus, 0.121 mol of NaOH reacts with same of HCl. Since, the molar mass of HCl is 34.46 g/mol, the given mass can be calculated as:

m = n X M = 0.121 X 36.46 = 4.421 g

Therefore, 4.421 g of HCl completely reacts with 4.85 g of NaCl.

(b) In the chemical equation of calcium hydroxide and nitric acid, the number of moles required by calcium hydroxide and nitric acid are:

$\text {HNO}_{3} +\text{Ca(OH)}_{2} \;\rightarrow\; 2 \text{H}_{2}\text{O}+ \text {Ca(NO)}_{3}}_{(2)}$

$\text {n}&=\frac{\text m}{\text M}\\\\\text {Molar Mass of Ca(OH)}_{2} = 74.093 g/mol\\\\\text {n}&=\frac{\text 4.85 }{\text 74.093}&=0.06545 \text{mol}$

Hence, 0.06545 mol of calcium hydroxide reacts completely with 0.1309 mol of nitric acid.

Now, the molar mass of nitric acid is 63.01 g/mol, thus, mass can be calculated as follows:

m = n X M = 0.1309 X 63.01 = 8.25 g

Therefore, 8.25 g of nitric acid completely reacts with

(c)The acid base neutralization balanced chemical reaction can be written as:

$\text {H}_{2} \text{SO}_{4} + 2 \text {KOH}\;\rightarrow\; 2 \text{H}_{2}\text{O}+ \text {K}_{2}\text {SO}_{4}$

The above reaction represents that for 2 mol of KOH, 1 mol of sulphuric acid reacts completely. The mass of KOH is 4.85g, converting into moles:

$\text {n}&=\frac{\text m}{\text M}\\\\\text {Molar Mass of KOH} = 56.105 g/mol\\\\\text {n}&=\frac{\text 4.85 }{\text 56.105}&=0.0864 \text{mol}$

Thus, 0.0864 mol of KOH reacts with 0.0432 mol of sulphuric acid.

Molar mass of sulphuric acid is 98.079 g/mol, the mass can be calculated as:

m = n X M = 0.0432 mol X 98.079 g/mol = 4.24 g

Therefore, 4.24 g of KOH completely reacts with 4.85 g of sulphuric acid.

To know more about neutralization reaction, refer to the following link:

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Answer from: pineapplepizaaaaa

Acid-base neutralization reaction is defined as reaction of acid with base to form corresponding salt and water. Strong acid and base completely neutralize each other.

(a) The acid base neutralization reaction is as follows:

$HCl(aq)+NaOH(aq)\rightarrow H_{2}O(l)+NaCl(g)$

From the above balanced chemical reaction, 1 mol of NaOH completely reacts with 1 mol of HCl. The mass of NaOH is given 4.85 g, convert this into number of moles as follows:

$n=\frac{m}{M}$

Molar mass of NaOH is 39.997 g/mol, thus,

$n=\frac{4.85 g}{39.997 g/mol}=0.121 mol$

Thus, 0.121 mol of NaOH reacts with same amount of HCl and number of moles of HCl will be 0.121 mol.

Since. molar mass of HCl is 36.46 g/mol, thus, mass can be calculated as follows:

$m=n\times M=0.121 mol\times 36.46 g/mol=4.421 g$

Therefore, 4.421 g of HCl completely reacts with 4.85 g of NaCl.

(b) The acid base neutralization reaction is as follows:

$2HNO_{3}(aq)+Ca(OH)_{2}(aq)\rightarrow 2H_{2}O(l)+Ca(NO_{3})_{2}(aq)$

From the above balanced chemical reaction, 1 mol of $Ca(OH)_{2}$ completely reacts with 2 mol of $HNO_{3}$ . The mass of $Ca(OH)_{2}$ is given 4.85 g, convert this into number of moles as follows:

$n=\frac{m}{M}$

Molar mass of $Ca(OH)_{2}$ is 74.093 g/mol, thus,

$n=\frac{4.85 g}{74.093 g/mol}=0.06545 mol$

Thus, 0.06545 mol of $Ca(OH)_{2}$ reacts with $2\times 0.06545 mol=0.13091 mol$ of $HNO_{3}$

Since. molar mass of $HNO_{3}$ is 63.01 g/mol, thus, mass can be calculated as follows:

$m=n\times M=0.13091 mol\times 63.01 g/mol=8.25 g$

Therefore, 8.25 g of $HNO_{3}$ completely reacts with 4.85 g of $Ca(OH)_{2}$ .

$H_{2}SO_{4}(aq)+2 KOH (aq)\rightarrow 2H_{2}O(l)+K_{2}SO_{4}(aq)$

From the above balanced chemical reaction, 2 mol of KOH completely reacts with 1 mol of $H_{2}SO_{4}$ . The mass of KOH is given 4.85 g, convert this into number of moles as follows:

$n=\frac{m}{M}$

Molar mass of KOH is 56.1056 g/mol, thus,

$n=\frac{4.85 g}{56.1056 g/mol}=0.0864 mol$

Thus, 0.0864 mol of KOH reacts with $\frac{0.0864 mol}{2}=0.0432 mol$ of $H_{2}SO_{4}$

Since. molar mass of $H_{2}SO_{4}$ is 98.079 g/mol, thus, mass can be calculated as follows:

$m=n\times M=0.0432 mol\times 98.079 g/mol=4.24 g$

Therefore, 4.24 g of KOH completely reacts with 4.85 g of $H_{2}SO_{4}$ .

Answer from: Quest

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Answer from: Quest

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