For the following reaction, 5.04 grams of nitrogen gas are allowed to react with 8.98 grams of oxygen gas: nitrogen(g) + oxygen(g) --> nitrogen monoxide(g)

a. What is the maximum mass of nitrogen monoxide that can be formed?

b. What is the FORMULA for the limiting reagent?

c. What mass of the excess reagent remains after the reaction is complete

1. 10.8 g of NO

2. N₂ is the limting reagent

3. 3.2 g of O₂ does not react

Explanation:

We determine the reaction: N₂(g) + O₂(g) →  2NO(g)

We need to determine the limiting reactant, but first we need the moles of each:

5.04 g / 29 g/mol = 0.180 moles N₂

8.98 g / 32 g/mol = 0.280 moles O₂

Ratio is 1:1, so the limiting reactant is the N₂. For 0.280 moles of O₂ I need the same amount, but I only have 0.180 moles of N₂

Ratio is 1:2. 1 mol of N₂ can produce 2 moles of NO

Then, 0.180 moles of N₂ may produce (0.180 .2) / 1 =  0.360 moles NO

If we convert them to mass → 0.360 mol . 30 g/1 mol = 10.8 g

As ratio is 1:1, for 0.180 moles of N₂, I need 0.180 moles of O₂.

As I have 0.280 moles of O₂, (0.280 - 0.180 ) = 0.100 moles does not react.

0.1 moles . 32 g/mol = 3.2 g of O₂ remains after the reaction is complete.

pie chart because it is in percents ( % )

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.