Answer:
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.
