Answer:
3.64g
Explanation:
Given parameters:
Mass of NH₃ = 18.1g
Mass of Cu₂O = 90.4g
Unknown:
Limiting reactant = ?
Mass of N₂ formed = ?
Solution:
The reaction equation is given as:
Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O
The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;
Number of moles =
Molar mass of Cu₂O = 2(63.6) + 16 = 143.2g/mol
Molar mass of NH₃ = 14 + 3(1) = 17g/mol
Number of moles of Cu₂O = = 0.13moles
Number of moles of NH₃ = = 5.32moles
From this reaction;
1 mole of Cu₂O combines with 2 mole of NH₃
So 0.13moles of Cu₂O will combine with 0.13 x 2 mole of NH₃
= 0.26moles of NH₃
Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;
Mass of N₂;
Mass = number of moles x molar mass
1 mole of Cu₂O will produce 1 mole of N₂
0.13 mole of Cu₂O will produce 0.13 mole of N₂
Mass = 0.13 x (2 x 14) = 3.64g
33233728793278237876548742787874578378572098-2932-=93788784787489
According to the question, the determined melting point of the compound is 112.5-113.0oC. When the solidified compound was retried, the melting point was found to be 133.6-154.5oC. This greater range higher than 112°C is caused by reusing samples leads to errors.
A pure sample is known by its sharp melting point. A pure sample does not melt over a large range. We can see this in the predetermined melting points of the pure sample(112.5-113.0oC).
However, reusing a sample introduces errors because the pure sample may become contaminated leading to a larger and higher range of melting point (133.6-154.5oC) which is far above 112°C.
Learn more: brainly.com/question/5325004
Glucose is extracted in the urine and all of the glucose is reabsorbed.
The balanced reaction equation for the combustion of butane is as follows;
C₄H₁₀ + 13/2O₂ ---> 4CO₂ + 5H₂O
the limiting reactant in this reaction is C₄H₁₀ This means that all the butane moles are consumed and amount of product formed depends on the amount of C₄H₁₀ used up.
stoichiometry of C₄H₁₀ to H₂O is 1:5
mass of butane used - 6.97 g
number of moles - 6.97 g / 58 g/mol = 0.12 mol
then the number of water moles produced - 0.12 mol x 5 = 0.6 mol
Therefore mass of water produced - 0.6 mol x 18 g/mol = 10.8 g