Whats the balance to Sr + O, SrO

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

Will mark BRAINLIEST question and choices in photos

ohaa [14]

Answer:

what to do in this question

8 0

3 years ago

Calculate the electrical energy per gram of anode material for the following reaction at 298 K:

antiseptic1488 [7]

The answer is:

E per gram = 0.45 V

The explanation:

when MnO2 is the substance who oxidized here so, the oxidizing agent and the anode here is Li.

and when the molar mass of Li is = 7 g/mol

and in our reaction equation we have 1 mole of Li will give 3.15 V of the electrical energy

that means that :

7 g of Li gives → 3.15 V

So 1 g of Li will give→ ???

∴ The E per gram = 3.15 V / 7 g of Li

= 0.45 V

6 0

3 years ago

Read 2 more answers

The following neutral atoms each have 8 places for electrons in their outermost shells.

Lunna [17]

C. ne with 8 valenc3 electrons

7 0

3 years ago

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In a sample of liquid water (H2O), which property differs among the water molecules?

garik1379 [7]

In a sample liquid water, a property that differs among the water molecules is its own orientation and space. The explanation of the answer to the question is because the number of the oxygen and hydrogen atoms and the hydrogen and atoms’ arrangement will never change and will always be the same for each of the water molecule. The strength of the bond of it also does not change and stays the same.

3 0

3 years ago

Whats the balance to Sr + O, SrO​

Whats the balance to Sr + O, SrO