55.36 is the mass of Calcium Bromide is needed to make 0.500 L of 0.554 M solution.
Explanation:
Data given:
mass of Ca = ?
molarity of the calcium bromide solution = 0.554 M
Volume of the calcium bromide solution= 0.5 L
First the moles of calcium bromide is calculated as:
molarity =
number of moles = 0.554 x 0.5
number of moles = 0.277 moles of calcium bromide
atomic mass of calcium bromide = 199.89 grams/mole
mass = atomic mass x number of moles
mass of calcium bromide = 199.89 x 0.277
= 55.36 grams
mass of calcium bromide required is 55.36 grams to make solution of 0.554 M in a solution of 0.5 litres.
Answer:
514.5 g.
Explanation:
- The balanced equation of the reaction is: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.
- It is clear that every 2.0 moles of NaOH react with 1.0 mole of H₂SO₄ to produce 1.0 mole of Na₂SO₄ and 2.0 moles of 2H₂O.
- Since NaOH is in excess, so H₂SO₄ is the limiting reactant.
- We need to calculate the no. of moles of 355.0 g of H₂SO₄:
n of H₂SO₄ = mass/molar mass = (355.0 g)/(98.0 g/mol) = 3.622 mol.
Using cross multiplication:
∵ 1.0 mol H₂SO₄ produces → 1.0 mol of Na₂SO₄.
∴ 3.622 mol H₂SO₄ produces → 3.662 mol of Na₂SO₄.
- Now, we can get the theoretical mass of Na₂SO₄:
∴ mass of Na₂SO₄ = no. of moles x molar mass = (3.662 mol)(142.04 g/mol) = 514.5 g.
Answer:
CaCl
2
Explanation:
As with all these problems, we assume a
100
⋅
g
of compound, and divide through by the ATOMIC masses of each component element:
Moles of calcium
=
36.1
⋅
g
40.08
⋅
g
⋅
mol
−
1
=
0.901
⋅
mol
.
Moles of chlorine
=
63.9
⋅
g
35.45
⋅
g
⋅
mol
−
1
=
1.802
mol
.
We divide each molar quantity through by the smallest molar quantity (that of calcium) to give an empirical formula of
CaCl
2
.
For ionic materials, we do not speak of the molecular formula, and the empirical formula is the formula we quote for reference.
helloo there the answer is Acetate