Answer:
2Li + F₂ → 2LiF
Explanation:
The reaction expression is given as:
Li + F₂ → LiF
We are to balance the expression. In that case, the number of atoms on both sides of the expression must be the same.
Let use a mathematical approach to solve this problem;
Assign variables a,b and c as the coefficients that will balance the expression:
aLi + bF₂ → cLiF
Conserving Li: a = c
F: 2b = c
let a = 1, c = 1 and b =
Multiply through by 2;
a = 2, b = 1 and c = 2
2Li + F₂ → 2LiF
Answer:
Ammonia is limiting reactant
Amount of oxygen left = 0.035 mol
Explanation:
Masa of ammonia = 2.00 g
Mass of oxygen = 4.00 g
Which is limiting reactant = ?
Balance chemical equation:
4NH₃ + 3O₂ → 2N₂ + 6H₂O
Number of moles of ammonia:
Number of moles = mass/molar mass
Number of moles = 2.00 g/ 17 g/mol
Number of moles = 0.12 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 4.00 g/ 32 g/mol
Number of moles = 0.125 mol
Now we will compare the moles of ammonia and oxygen with water and nitrogen.
NH₃ : N₂
4 : 2
0.12 : 2/4×0.12 = 0.06
NH₃ : H₂O
4 : 6
0.12 : 6/4×0.12 = 0.18
O₂ : N₂
3 : 2
0.125 : 2/3×0.125 = 0.08
O₂ : H₂O
3 : 6
0.125 : 6/3×0.125 = 0.25
The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.
Amount of oxygen left:
NH₃ : O₂
4 : 3
0.12 : 3/4×0.12= 0.09
Amount of oxygen react = 0.09 mol
Amount of oxygen left = 0.125 - 0.09 = 0.035 mol
Answer:
m= 29.645 g
Explanation:
Density:
Density is equal to the mass of substance divided by its volume.
Units:
SI unit of density is Kg/m3.
Other units are given below,
g/cm3, g/mL , kg/L
Formula:
D=m/v
D= density
m=mass
V=volume
Symbol:
The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.
Given data:
density of wood = 0.77 g/cm³
volume= 38.5 cm³
mass= ?
Solution:
d= m/v
m= d × v
m= 0.77 g/cm³× 38.5 cm³
m= 29.645 g
Answer:
number of moles = 0.21120811
Explanation:
To find the number of moles, given the mass of the solute, we use the formula:




Label the variables with the numbers in the problem:



The first thing we have to do is find the molar mass of sodium sulfate, in order for us to use the formula for finding the number of moles:
Formula for finding the molar mass of sodium sulfate:

For the variables and what they mean are below for finding the molar mass of sodium sulfate:





Plug the numbers into the formula, to find the molar mass of sodium sulfate:











Now that we have found the molar mass, we can calculate the number of moles in the solution of sodium sulfate with the formula:








0.21120811 rounded gives you 0.2112
or if you did the problem without decimals
30 grams of sodium sulfate divided by its molecular weight – which we found to be 142 – gives us a value of 0.2113 moles.