Answered

Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Get expert answers to your questions quickly and accurately from our dedicated community of professionals. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

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

Sagot :

anfabba15

Answer:

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

We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.