Molar Volume is required to solve this problem. As we know that "1 mole of any gas at standard temperature and pressure occupies 22.4 L of volume". SO using this concept, we can calculate the volume of ammonia formed by reacting 54.1 L of Hydrogen gas as follow,
<span>The choices are as follows:
h2o + 2o2 = h2o2
fe2o3 + 3h2 = 2fe + 3h2o
al + 3br2 = albr3
caco3 = </span><span>cao + co2
The correct answers would be the second and the last option. The equations that are correctly balanced are:
</span> fe2o3 + 3h2 = 2fe + 3h2o
caco3 = cao + co2
To balance, it should be that the number of atoms of each element in the reactant and the product side is equal.
Nothing happens to these particles when it comes to size however if it were to be speed, the sample would increase.
Fact: The size of a particle will never change!
The heat released by cpmolete the combustion of organic products with oxygen is called heat of combustion.
Here 1 mol of CH4 realesed 802.3 KJ
To emit 264 kJ you multiply you need (1 mol of CH4/802.3 kJ)* 264 kJ = 0.329 mol of CH4
The molar mass, MM, of CH4 is 12 g/mol + 4*1g/mol = 16 g/mol
The to obtain the mass multiply the number of moles times the molar mass:
mass = n * MM = 0.329mol * 16g/mol = 5.26 grams
Answer: 5.26 grams
