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True [87]
3 years ago
5

For each of the acid–base reactions, calculate the mass (in grams) of each acid necessary to completely react with and neutraliz

e 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)
Chemistry
1 answer:
LiRa [457]3 years ago
8 0

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}.

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

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}.

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<u>We are given:</u>

251 mL sample of 0.45M HCl added to 455 mL distilled water

<u>Whack a mole! (finding the number of moles):</u>

We know that in order to find molarity, we use the formula:

Molarity = number of moles / Volume (in L)

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Number of moles = Molarity * Volume(in L)

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<u>Time to concentrate (finding the final concentration):</u>

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___________________________________________________________

<u>BONUS METHOD TIME!!!</u>

We know the relation:

M1 * V1 = M2 * V2

where M1 and M2 are the initial and final molarities and V1 and V2 are initial and final volumes respectively

notice that I didn't mention that the volume has to be in Liters, that's because of the units being concerned with both sides of the equation, say I have the volume in mL and want to convert both these volumes to L, I would divide both sides by 1000, which would NOT change the overall value

Now, plugging values in this equation

(0.45) * (251) = (251 + 455)* (M2)

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