Answer:
0.544 M
Explanation:
First find the moles in the final solution
0.8 mols/L *1.7L
1.36 mols
so there is 1.36 mols in 2.5L
concentration will be 1.36/2.5
0.544 M
Answer:
H₂O is the limiting reactant
Theoretical yield of 240 g Al₂O₃ and 14 g H₂
Explanation:
Find how many moles of one reactant is needed to completely react with the other.
6.5 mol Al × (3 mol H₂O / 2 mol Al) = 9.75 mol H₂O
We need 9.75 mol of H₂O to completely react with 6.5 mol of Al. But we only have 7.2 mol of H₂O. Therefore, H₂O is the limiting reactant.
Now find the theoretical yield:
7.2 mol H₂O × (1 mol Al₂O₃ / 3 mol H₂O) × (102 g Al₂O₃ / mol Al₂O₃) ≈ 240 g Al₂O₃
7.2 mol H₂O × (3 mol H₂ / 3 mol H₂O) × (2 g H₂ / mol H₂) ≈ 14 g H₂
Since the data was given to two significant figures, we must round our answer to two significant figures as well.
Answer:
He's flying and was looking
Explanation:
Carbon is found in oil and gas.
Aluminum a light metal used in making pots and pans.
Bromine is used in photography.
Explanation:
We will balance equation which describes the reaction between sulfuric acid and sodium bicarbonate: as follows.
Next we will calculate how many moles of
are present in 85.00 mL of 1.500 M sulfuric acid.
As, Molarity = 
1.500 M = 
n = 0.1275 mol
Now set up and solve a stoichiometric conversion from moles of
to grams of
. As, the molar mass of
is 84.01 g/mol.
= 21.42 g
So unfortunately, 15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.