<h3>
Answer:</h3>
18.58 liters of hydrogen gas
<h3>
Explanation:</h3>
We are given;
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
is 57.1 g and we are asked to find number of moles present in 57.1 g of