Question

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179 x 1023 atoms of magnesium with 54.21 g of phosphoric acid (H3PO4)?
3 Mg + 2 H3(PO4) --> Mg3(PO4)2 + 3 H2

Answer 1

Answer:

18.58 liters of hydrogen gas

Explanation:

We are given;

• The equation;

3Mg + 2H₃(PO₄) → Mg₃(PO₄)₂ + 3H₂

• Atoms of Magnesium = 7.179 x 10^23 atoms
• Mass of phosphoric acid as 54.21 g

We are required to determine the volume of hydrogen gas produced;

Step 1; moles of Magnesium

1 mole of an element contains 6.02 × 10^23 atoms

therefore;

Moles of Mg = (7.179 x 10^23 ) ÷ (6.02 × 10^23)

                   = 1.193 moles

Step 2: Moles of phosphoric acid

moles = Mass ÷ Molar mass

Molar mass of phosphoric acid = 97.994 g/mol

Therefore;

Moles of Phosphoric acid = 54.21 g ÷ 97.994 g/mol

                                           = 0.553 moles

Step 3: Determine the rate limiting reagent

From the mole ratio of Mg to Phosphoric acid (3 : 2);

1.193 moles of magnesium requires 0.795 moles of phosphoric acid while,

0.0553 moles of phosphoric acid requires 0.8295 moles of Mg

Therefore, phosphoric acid is the rate limiting reagent

step 4: Determine the moles of hydrogen produced

From the equation, w moles of phosphoric acid reacts to produce 3 moles of hydrogen;

Therefore; moles of Hydrogen = moles of phosphoric acid × 3/2

                                                   = 0.553 moles × 3/2

                                                   = 0.8295 moles

Step 5: Volume of hydrogen gas

1 mole of a gas occupies a volume of 22.4 liters at STP

Therefore;

Volume of Hydrogen = 0.8295 moles × 22.4 L/mol

                                  = 18.58 Liters

Therefore; 18.58 liters of hydrogen gas  will be produced