What volume of hydrogen gas at STP will be produced from 36.7 g of magnesium reacting with an excess amount of hydrochloric acid?

V = 34.3 L

Explanation:

Given data:

Volume of hydrogen produced = ?

Mass of magnesium = 36.7 g

Temperature and pressure = standard

Solution:

Mg + 2HCl       →   MgCl₂ + H₂

Number of moles of magnesium:

Number of moles = mass/molar mass

Number of moles = 36.7 g/ 24 g/mol

Number of moles = 1.53 mol

Now we will compare the moles of magnesium with hydrogen.

                         Mg           :          H₂

                          1              :           1

                         1.53          :          1.53

Number of moles of hydrogen produced = 1.53 mol

Volume of hydrogen:

PV = nRT

V = nRT / P

V = 1.53 mol × 0.0821 atm.L/mol.K × 273 K/ 1 atm

V = 34.3 L

electron microscopy and scanning tunneling microscopes gives us a greater understanding of atoms by being able to see the atom itself, and lets us see some atomic reactions.

explanation:

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.