How many liters of hydrogen gas will be produced at STP from the reaction of 5.159 x 1021 atoms of magnesium with 55.23 g of phosphoric acid (H3PO4)? 3 Mg + 2 H3(PO4) Mg3(PO4)2 + 3 H2

0.190L of hydrogen may be produced by the reaction.

Explanation:

Our reaction is:

3Mg + 2H₃PO₄ →  Mg₃(PO₄)₂ + 3H₂

We need to determine the limting reactant. Let's find out the moles of each:

5.159×10²¹ atoms . 1 mol / 6.02×10²³ atoms = 0.00857 moles of Mg

55.23 g . 1 mol / 97.97 g = 0.563 moles of acid

2 moles of acid react to 3 moles of Mg

0.563 moles of acid may react to: (0.563 . 3) /2 = 0.8445 moles of Mg

Definetely the limting reactant is Mg.

As ratio is 3:3, 3 moles of Mg can produce 3 moles of hydrogen

Then, 0.00857 moles of Mg must produce 0.00857 moles of H₂

At STP, 1 mol of any gas occupies 22.4L

0.00857 mol . 22.4L / 1mol = 0.190L

b.

the accuracy of the data.