Consider the combustion of octane: 2C8H18(g) + 2502(g) ===> 16CO2(g) + 18H2O(g)
Calculate the mass of water (H20) produced from 11.2 grams of octane (C3H18) assuming excess
oxygen.
15.9 g
O 0.196 g
O 4.31 g
1.77g
14.1g

15.9 g

Explanation:

(Take the atomic mass of C=12.0, H=1.0, O=16.0)

no. of moles = mass / molar mass

no. of moles of octane used = 11.2 / (12.0x8 + 1x18)

= 0.0982456 mol

Since oxygen is in excess and octane is the limiting reagent, the no. of moles of H2O depends on the no. of moles of octane used.

From the balanced equation, the mole ratio of octane : water = 2:18 = 1: 9,

so this means, one mole of octane produced 9 moles of water.

Using this ratio, we can deduce that (y is the no. of moles of water produced):

y = 0.0982456x9

y= 0.88421 mol

Since mass = no. of moles x molar mass,

mass of water produced = 0.88421  x (1.0x2+16.0)

=15.9 g

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.  

it is reliable if two scientists recorded it