If the acetylene tank contains 37.0 mil of C2H2 and the oxygen tank contains 81.0 mol of O2, what is the limiting reactant for this reaction
2C2H2(l)+502(g) 4CO2(g)+2H2O(g)

Explanation:

2C₂H₂(l)  +  50₂(g) =   4CO₂(g)  +  2H₂O(g)

2 mol          5 mol

2 mol of C₂H₂ reacts with 5 mol of O₂

37 mol of C₂H₂  will require 5/2 x 37 mol of O₂

= 92.5 mol of O₂

But oxygen available is only 81 mol so formation of product will be controlled by oxygen available . Hence oxygen is the limiting factor.

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.