You're looking for the number of moles of H2, and you have 6.0 mol Al and 13 mol HCL.
For the first part, you have to make your way from 6.0 mol of Al to mol of H2, right? For that to happen, you need to make a conversion factor that will cancel the mol Al, in such case use the 2 moles of Al from your equation to cancel them out. At the top of the equation, you can use the number of moles of H2 from the equation and find the moles that will be produced for the H2.
6.0mol Al x 3 mol H2/2 mol Al = 9 mol H2
For the second part, you have to make the same procedure, make a conversion factor that will cancel the mol of HCL and for that you need to use the 6 mol HCL from your equation, and at the numerator you can put the 3 mol of H2 from the equation so that you can find the number of moles of H2 that will be produced.
13 mol HCL x 3 mol H2/6 mol HCL = 6.5 mol H2
As it can be seen, HCL produces the less amount of H2 moles. Therefore, the reaction CANNOT produce more than 6.5 mol H2, in that case 6.5 mol will be the maximum number of moles that will be produced at the end because HCL does not have enough to produce more than 6.5 mol.
In that case HCL is the limiting reactant because it limits that will be produced, and so the answer is B!
Answer:
Empirical formula of C₈H₈ = CH
Explanation:
Data Given:
Molecular Formula = C₈H₈
Empirical Formula = ?
Solution
Empirical Formula:
Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.
So,
tha ration of the molecular formula should be divided by whole number to get the simplest ratio of molecule
C₈H₈ Consist of Carbon (C), and Hydrogen (H)
Now
Look at the ratio of these two atoms in the compound
C : H
8 : 8
Divide the ratio by two to get simplest ratio
C : H
8/8 : 8/8
1 : 1
So for the empirical formula is the simplest ratio of carbon to hydrogen 1 : 1
So the empirical formula will be
Empirical formula of C₈H₈ = CH
Well not calculus because that has nothing, well mostly nothing to do with balancing chemical equation, so B or C. Now for me personally B is way faster, though C is sometimes faster if you get lucky the way to solve it is B
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
