. Sulfuric acid reacts with calcium hydroxide to form calcium sulfate and water.

a. Write the balanced equation for the reaction.

b. Find the limiting reactant when 75.0 g sulfuric acid reacts with 55.0 g calcium hydroxide.

c. Find the moles of unreacted starting material

A. H2SO4 + Ca(OH)2 —> CaSO4 + 2H2O

B. Calcium hydroxide Ca(OH)2 is the limiting reactant.

C. The mole of the unreacted starting material is 0.022 mole.

Explanation:

A. The balanced equation for the reaction.

This is illustrated below:

H2SO4 + Ca(OH)2 —> CaSO4 + 2H2O

B. Determination oh the limiting reactant. The limiting reactant reactant can be obtained as follow:

Step 1:

Determination of the masses of H2SO4 and Ca(OH)2 that reacted from the balanced equation. This is illustrated below:

Molar Mass of H2SO4 = (2x1) + 32 + (16x4) = 2 + 32 + 64 = 98g/mol

Molar Mass of Ca(OH)2 = 40 + 2(16 + 1) = 40 + 2(17) = 40 + 34 = 74g/mol.

Thus, from the balanced equation above,

98g of H2SO4 reacted with 74g of Ca(OH)2.

Step 2:

Determination of the limiting reactant. The limiting reactant can be obtained as follow:

From the balanced equation above,

98g of H2SO4 reacted with 74g of Ca(OH)2.

Therefore, 75g of H2SO4 will react with = (75 x 74)/98 = 56.63g of Ca(OH)2.

From the calculations made above, we discovered that it will take a higher mass of Ca(OH)2 i.e 56.63g than what was given i.e 55g to react with 75g of H2SO4.

Therfore, Ca(OH)2 is the limiting reactant.

C. Determination of the moles of unreacted starting material.

The excess reactant is H2SO4.

The unreacted moles of H2SO4 can be obtained as follow:

Step 1:

Determination of the mass of H2SO4 needed to react with 55g of Ca(OH)2. This is illustrated below:

From the balanced equation above,

98g of H2SO4 reacted with 74g of Ca(OH)2.

Therefore, Xg of H2SO4 will react with 55g of Ca(OH)2 i.e

Xg of H2SO4 = (98 x 55)/74

Xg of H2SO4 = 72.84g

Therefore, 72.84g of H2SO4 reacted.

Step 2:

Determination of the unreacted Mass of H2SO4. This is shown below:

Mass of H2SO4 given = 75g

Mass of H2SO4 that reacted = 72.84g

Mass of unreacted H2SO4 =..?

Mass of unreacted H2SO4 = (Mass of H2SO4 given) – (Mass of H2SO4 that reacted)

Mass of unreacted H2SO4 = 75 – 72.84

Mass of unreacted H2SO4 = 2.16g

Step 3:

Conversion of the unreacted mass of H2SO4 to mole. This is illustrated below:

Mass of unreacted H2SO4 = 2.16g

Molar Mass of H2SO4 = (2x1) + 32 + (16x4) = 2 + 32 + 64 = 98g/mol

Number of mole of unreacted H2SO4 =?

Number of mole = Mass/Molar Mass

Number of mole of unreacted H2SO4 = 2.16/98 = 0.022 mole.

Therefore, the mole of the unreacted starting material is 0.022 mole.

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.