Question

What volume of hydrogen gas will be formed at STP from the reaction of 1.30g of aluminum with 120.00mL of 0.150M sulfuric acid?

How do you make 150.0 mL of 0.250 M NH3 from a stock bottle that is 10.0 M?

Answer 1

1. The balanced equation is written as

2Al(s)+3H₂SO₄(aq) ? Al₂(SO₄)₃(aq)+3H₂(g)

   Molarity of H2SO4 = 0.150 M, Volume of H2SO4 = 120 x 10^-3 L

   Number of moles of H2SO4 given = 0.150 x 120 x 10^-3 = 0.018 moles

From equation we can see that 3 moles of H2SO4 produce 3 moles of H2 gas

Hence we can say 1 mole of H2SO4 produce 1 mole of H2 gas

Hence 0.018 moles of H2SO4 produce 0.018 moles of H2 gas

Therefore number of moles of H2 gas = weight of H2 gas/ Molecular weight of H2 gas

0.018 moles = weight of H2 gas/ 2

weight of H2 gas = 2 x 0.018 = 0.036 g

Now 1 mole of any gas at STP occupies 22.4 liters of volume

therefore 0.018 moles of H2 gas occupies = 0.018 x 22.4 liters of volume

= 0.4032 L = 403.2 mL

2. M1 V1 = M2 V2

0.250 x 0.15 = 10.0 x V2

3.75 x 10^-3 L = V2

3.75 mL = V2

   Since we are making a solution that is approximately 0.250 M NH3 we can use a graduated cylinder to measure the 3.75 mL of concentrated NH3, transfer the NH3 to a beaker, and add sufficient water to give a total volume of approximately 150 mL.