<span> Mg(OH)2(s) + 2HCl(aq) yield MgCl2(aq) + 2H2O(l)
grams HCl required = (50.6 grams Mg(OH)2) * (1 mol Mg(OH)2 / 58.3197 grams Mg(OH)2) * (2 mol HCl / 1 mol Mg(OH)2) * (36.453 grams HCl / 1 mol HCl) = 63.26 grams HCl required
Since there are only 45.0 grams HCl, then HCl is the limiting reactant.
theoretical yield MgCl2 = (45.0 grams HCl) * (1 mol HCl / 36.453 grams HCl) * (1 mol MgCl2 / 2 mol HCl) * (95.211 grams MgCl2 / 1 mol MgCl2) = 58.6 grams MgCl2 </span>
The outer shell electrons are only involved in the bonding process since they are the only 'incomplete' shell and it needs to be fulfilled by another element.
Answer:
S = 21.92 %
F = 78.08 %
Explanation:
To find the percent composition of each element in SF6, we must find the molar mass of SF6 first.
Molar mass of SF6 = 32 + 19(6)
= 32 + 114
= 146g/mol
mass of Sulphur (S) in SF6 = 32g
mass of Fluorine (F) in SF6 = 114g
Percent composition = mass of element/molar mass of compound × 100
- % composition of S = 32/146 × 100 = 21.92%.
- % composition of F = 114/146 × 100 = 78.08%.
Answer:
Explanation:
This is a limiting reactant problem.
Mg(s)
+
2HCl(aq)
→
MgCl
2
(
aq
)
+ H
2
(
g
)
Determine Moles of Magnesium
Divide the given mass of magnesium by its molar mass (atomic weight on periodic table in g/mol).
4.86
g Mg
×
1
mol Mg
24.3050
g Mg
=
0.200 mol Mg
Determine Moles of 2M Hydrochloric Acid
Convert
100 cm
3
to
100 mL
and then to
0.1 L
.
1 dm
3
=
1 L
Convert
2.00 mol/dm
3
to
2.00 mol/L
Multiply
0.1
L
times
2.00 mol/L
.
100
cm
3
×
1
mL
1
cm
3
×
1
L
1000
mL
=
0.1 L HCl
2.00 mol/dm
3
=
2.00 mol/L
0.1
L
×
2.00
mol
1
L
=
0.200 mol HCl
Multiply the moles of each reactant times the appropriate mole ratio from the balanced equation. Then multiply times the molar mass of hydrogen gas,
2.01588 g/mol
0.200
mol Mg
×
1
mol H
2
1
mol Mg
×
2.01588
g H
2
1
mol H
2
=
0.403 g H
2
0.200
mol HCl
×
1
mol H
2
2
mol HCl
×
2.01588
g H
2
1
mol H
2
=
0.202 g H
2
The limiting reactant is
HCl
, which will produce
0.202 g H
2
under the stated conditions.
pls mark as brainliest ans
Answer:
V = 22.41 L
Explanation:
Given data:
Mass of nitrogen = 14.0 g
Volume of gas at STP = ?
Gas constant = 0.0821 atm.L/mol.K
Solution:
Number of moles of gas:
Number of moles = mass/molar mass
Number of moles= 14 g/ 14 g/mol
Number of moles = 1 mol
Volume of gas:
PV = nRT
1 atm × V = 1 mol × 0.0821 atm.L/mol.K × 273 K
V = 22.41 atm.L / 1 atm
V = 22.41 L
