Your answer is probably
Vaporization point
Empirical formula is the simplest ratio of components making up a compound.
The percentage composition of each element has been given
therefore the mass present of each element in 100 g of compound is
B N H
mass 40.28 g 52.20 g 7.53 g
number of moles
40.28 g / 11 g/mol 52.20 g / 14 g/mol 7.53 g / 1 g/mol
= 3.662 mol = 3.729 mol = 7.53 mol
divide the number of moles by the least number of moles, that is 3.662
3.662 / 3.662 3.729 / 3.662 7.53 / 3.662
= 1.000 = 1.018 = 2.056
the ratio of the elements after rounding off to the nearest whole number is
B : N : H = 1 : 1 : 2
therefore empirical formula for the compound is B₁N₁H₂
that can be written as BNH₂
Answer:
NO would form 65.7 g.
H₂O would form 59.13 g.
Explanation:
Given data:
Moles of NH₃ = 2.19
Moles of O₂ = 4.93
Mass of NO produced = ?
Mass of produced H₂O = ?
Solution:
First of all we will write the balance chemical equation,
4NH₃ + 5O₂ → 4NO + 6H₂O
Now we will compare the moles of NO and H₂O with ammonia from balanced chemical equation:
NH₃ : NO NH₃ : H₂O
4 : 4 4 : 6
2.19 : 2.19 2.19 : 6/4 × 2.19 = 3.285 mol
Now we will compare the moles of NO and H₂O with oxygen from balanced chemical equation:
O₂ : NO O₂ : H₂O
5 : 4 5 : 6
4.93 : 4/5×4.93 = 3.944 mol 4.93 : 6/5 × 4.93 = 5.916 mol
we can see that moles of water and nitrogen monoxide produced from the ammonia are less, so ammonia will be limiting reactant and will limit the product yield.
Mass of water = number of moles × molar mass
Mass of water = 3.285 mol × 18 g/mol
Mass of water = 59.13 g
Mass of nitrogen monoxide = number of moles × molar mass
Mass of nitrogen monoxide = 2.19 mol × 30 g/mol
Mass of nitrogen monoxide = 65.7 g
