2. Heat is:

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

LiRa [457]

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 $H_{2}SO_{4}$ .

8 0

3 years ago

You have 500,000 atoms of a radioactive substance. After 2 half-lives have past, how many atoms remain?

Aleonysh [2.5K]

Answer:

125000

Explanation:

Because it is halved and halved again.

3 0

2 years ago

Read 2 more answers

A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant?

KengaRu [80]

Answer:

d. Copper (II) sulfate

Explanation:

Given data:

Mass of Al = 1.25 g

Mass of CuSO₄ = 3.28 g

What is limiting reactant = ?

Solution:

Chemical equation:

2Al + 3CuSO₄ → Al₂ (SO₄)₃ + 3Cu

Number of moles of Al:

Number of moles = mass/molar mass

Number of moles = 1.25 g/ 27 g/mol

Number of moles = 0.05 mol

Number of moles of CuSO₄:

Number of moles = mass/molar mass

Number of moles = 3.28 g/ 159.6 g/mol

Number of moles = 0.02 mol

now we will compare the moles of reactant with product.

Al : Al₂ (SO₄)₃

2 : 1

0.05 : 1/2×0.05=0.025 mol

Al : Cu

2 : 3

0.05 : 3/2×0.05 = 0.075 mol

CuSO₄ : Al₂ (SO₄)₃

3 : 1

0.02 : 1/3×0.02=0.007 mol

CuSO₄ : Cu

3 : 3

0.02 : 0.02

Less number of moles of reactants are produced by CuSO₄ thus it will act as limiting reactant.

4 0

3 years ago

Plants contain xylem and phloem tissues for carrying water and food. What organ system in animals performs a similar function as

Nuetrik [128]

Answer:

digestive system may be the answer

3 0

3 years ago

5. Water, wind, ice, and gravity are important agents of ___ which is a destructive force

Julli [10]

I agree with the statement that the other person has made

6 0

2 years ago