Answer: 6.38 moles of 
are needed to react with 2.55 mol of 
Explanation:
The balanced chemical reaction for the combustion of ethyne is:
According to stoichiometry:
2 moles of require = 5 moles of
Thus 2.55 mol of require = 
moles of
Thus 6.38 moles of are needed to react with 2.55 mol of
The small part in the center of the atom is called a nucleus
A dipeptide is formed when 2 amino acids are joined together by a peptide linkage.
Each amino acid molecule contains an amino group ( -NH₂) and a carboxyl ( -COOH) group. During peptide formation, a carboxyl group of one amino acid reacts with amino group of the other , resulting in the formation of a linkage known as "peptide linkage". A water molecule is also eliminated in this reaction.
A general peptide formation reaction can be written as
In the above reaction, we can see a peptide linkage ( -CONH-) getting formed between 2 amino acid molecules.
In the given molecule the peptide linkage is shown by drawing a box around it. Please refer to the attached image.
If we want to know the amino acid molecules from which this peptide is formed, we will simply break the CO-NH bond.
We will add -OH group to CO part and -H to NH part to get 2 amino acid molecules as shown in the picture.
The two amino acids thus formed are Serine & Cysteine.
Serine is designated as "Ser" and Cysteine is designated as "Cys"
Therefore the given dipeptide is designated as "Ser-Cys"
Answer:
1) The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
2) The amount (in grams) of excess reactant H₂ = 4.39 g.
Explanation:
- Firstly, we should write the balanced equation of the reaction:
<em>N₂ + 3H₂ → 2NH₃.</em>
<em>1) To determine the limiting reactant of the reaction:</em>
- From the stichiometry of the balanced equation, 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.
- This means that <em>N₂ reacts with H₂ with a ratio of (1:3).</em>
- We need to calculate the no. of moles (n) of N₂ (5.23 g) and H₂ (5.52 g) using the relation:<em> n = mass / molar mass.</em>
The no. of moles of N₂ in (5.23 g) = mass / molar mass = (5.23 g) / (28.00 g/mol) = 0.1868 mol.
The no. of moles of H₂ (5.52 g) = mass / molar mass = (5.52 g) / (2.015 g/mol) = 2.74 mol.
- From the stichiometry, N₂ reacts with H₂ with a ratio of (1:3).
The ratio of the reactants of N₂ (5.23 g, 0.1868 mol) to H₂ (5.52 g, 2.74 mol) is (1:14.67).
∴ The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.
<em>2) To determine the amount (in grams) of excess reactant of the reaction:</em>
- As showed in the part 1, The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
- Also, 0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.
- The no. of moles are in excess of H₂ = 2.74 mol - 0.5604 mol (reacted with N₂) = 2.1796 mol.
- ∴ The amount (in grams) of excess reactant H₂ = n (excess moles) x molar mass = (2.1796 mol)((2.015 g/mol) = 4.39 g.
