Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
Answer:
A. It formed by barium(Ba+2) ion and sulfate ( SO42- )
B. It is formed by calcium ion (Ca+2) and two fluoride ions (2F-)
C. It is formed by magnesium ion (Mg+2) and nitride ion (N3-)
D. It is formed by two potassium ions (2 K+) and oxide ion(O2-)
Answer: D. 19.9 g hydrogen remains.
Explanation:
To calculate the moles, we use the equation:
a) moles of 
b) moles of 
According to stoichiometry :
1 mole of require 1 mole of
Thus 0.0787 moles of 
require=
of
Thus is the limiting reagent as it limits the formation of product and acts as the excess reagent. (10.0-0.0787)= 9.92 moles of are left unreacted.
Mass of 
Thus 19.9 g of remains unreacted.
Answer:
If the half-life of 14C is 5730 years, when this period of time has passed it will have been halved, it is called the exponential decay law of radioactive isotopes.
