Determine the volume (mL) of 15.0 M sulfuric acid needed to react with 45.0 g of aluminum to produce aluminum sulfate.

167 mL.

Explanation:

We'll begin by calculating the number of moles in 45 g of aluminum (Al). This can be obtained as follow:

Mass of Al = 45 g

Molar mass of Al = 27 g/mol

Mole of Al =?

Mole = mass /Molar mass

Mole of Al = 45/27

Mole of Al = 1.67 moles

Next, the balanced equation for the reaction. This is given below:

2Al + 3H2SO4 → Al2(SO4)3 + 3H2

From the balanced equation above,

2 moles of Al reacted with 3 moles of H2SO4.

Next, we shall determine the number of mole of H2SO4 needed to react with 45 g (i.e 1.67 moles) of Al. This can be obtained as:

From the balanced equation above,

2 moles of Al reacted with 3 moles of H2SO4.

Therefore, 1.67 moles of Al will react with = (1.67 × 3)/2 = 2.505 moles of H2SO4.

Thus 2.505 moles of H2SO4 is needed for the reaction.

Next, we shall determine the volume of H2SO4 needed for the reaction. This can be obtained as follow:

Molarity of H2SO4 = 15.0 M

Mole of H2SO4 = 2.505 moles

Volume =?

Molarity = mole /Volume

15 = 2.505 / volume

Cross multiply

15 × volume = 2.505

Divide both side by 15

Volume = 2.505/15

Volume = 0.167 L

Finally, we shall convert 0.167 L to mL. This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.167 L = 0.167 L × 1000 mL / 1 L

0.167 L = 167 mL

Thus, 0.167 L is equivalent to 167 mL.

Therefore, 167 mL H2SO4 is needed for the reaction.

i'm assuming that the correct answer is indeed the first one/ "a"..