Answer:
3.1 kg
Explanation:
Step 1: Write the balanced combustion equation
C₈H₁₈ + 12.5 O₂ ⇒ 8 CO₂ + 9 H₂O
Step 2: Calculate the moles corresponding to 1.0 kg of C₈H₁₈.
The molar mass of C₈H₁₈ is 114.23 g/mol.
1.0 × 10³ g × 1 mol/114.23 g = 8.8 mol
Step 3: Calculate the moles of CO₂ produced from 8.8 moles of C₈H₁₈
The molar ratio of C₈H₁₈ to CO₂ is 1:8. The moles of CO₂ produced are 8/1 × 8.8 mol = 70 mol.
Step 4: Calculate the mass corresponding to 70 moles of CO₂
The molar mass of CO₂ is 44.01 g/mol.
70 mol × 44.01 g/mol = 3.1 × 10³ g = 3.1 kg
Answer:
A chemical equation can be described as a symbolic representation of a chemical reaction. It can be written in the form of words or symbols.
The following chemical reaction can be completed as :
Na2SO4·10H2O + 2C = Na2S + 10H2O + 2CO2
Sodium Sulfate Decahydrate + Diamond = Sodium Sulfide + Water + Carbon Dioxide
The products of the reaction are:
- Sodium Sulfide (Na2S)
- Water (H2O)
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.
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Answer:
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