Answer:
a. 4.41 g of Urea
b. 1.5 g of Urea
Explanation:
To start the problem, we define the reaction:
2NH₃ (g) + CO₂ (g) → CH₄N₂O (s) + H₂O(l)
We only have mass of ammonia, so we assume the carbon dioxide is in excess and ammonia is the limiting reactant:
2.6 g . 1mol / 17g = 0.153 moles of ammonia
Ratio is 2:1. 2 moles of ammonia can produce 1 mol of urea
0.153 moles ammonia may produce, the half of moles
0153 /2 = 0.076 moles of urea
To state the theoretical yield we convert moles to mass:
0.076 mol . 58 g/mol = 4.41 g
That's the 100 % yield reaction
If the percent yield, was 34%:
4.41 g . 0.34 = 1.50 g of urea were produced.
Formula is (Yield produced / Theoretical yield) . 100 → Percent yield
Answer:
Sr would be the limiting reactant
5 moles
Explanation:
Since the equation is a balanced equation, the coefficient shows how each substance relates to the other in terms of the number of moles.
Reactants would be those on the left hand side of the arrow, while the products would be found on te right and side of the arrow. In this question, the reactants would be Sr and O₂.
Limiting reactant is the reactant that is insufficient; meaning to say that there is not enough of that substance and thus the reaction cannot continue. The other reactant(s) that is not limiting is called the excess reactants.
From the balanced equation, 2 moles of Sr is needed to react with 1 mole of O₂. Thus, if we have 5 moles of each reactant, Sr would be the limiting reactant since for every 1 mole of O₂, there has to be 2 moles of Sr in order for the reaction to proceed. Thus, if we have 5 moles of O₂, we would need 10 moles of Sr.
When we work out the amount of products formed, we look at the number of moles of the limiting reactant. This is because the limiting reactant determines how much is being reacted, while the excess number of moles of the excess reactant will remain unreacted.
For every 2 moles of Sr reacted, 2 moles of SrO would be produced. This means that the mole ratio of Sr to SrO is 1:1. Thus, since 5 moles of Sr has been reacted, 5 moles of the product (SrO) would be produced.
A pan on the stove getting hot, because the pan is conducting the heat from the stove.
0.600
D) 0.600 is the final concentration of the solution of KCl.
V1 = 50.0 mL.
