Answer is: <span>the mass of the excess reactant (ethane) leftover is 90.135 grams.
</span>Chemical reaction: 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O<span>(g).
m(</span>C₂H₆) = 152 g.
n(C₂H₆) = m(C₂H₆) ÷ M(C₂H₆).
n(C₂H₆) = 152 g ÷ 30 g/mol.
n(C₂H₆) = 5.067 mol.
m(O₂) = 231 g.
n(O₂) = 231 g ÷ 32 g/mol.
n(O₂) = 7.218 mol; limiting reactant.
From chemical reaction: n(O₂) : n(C₂H₆) = 7 : 2.
n(C₂H₆) = 2 · 7.218 mol ÷ 7.
n(C₂H₆) = 2.0625mol.
Δn(C₂H₆) = 5.067 mol - 2.0625 mol.
Δn(C₂H₆) = 3.0045 mol.
Δm(C₂H₆) = 3.0045 mol · 30 g/mol = 90.135 g.
Answer:
The answer to your question is V2 = 1.3 l
Explanation:
Data
Pressure 1 = P1 = 1.5 atm
Volume 1 = V1 = 2.50 l
Temperature 1 = T1 = 22°C
Pressure 2 = P2 = 3 atm
Volume 2 = V2 = ?
Temperature 2 = T2 = 35°C
Process
1.- Convert temperature to °K
T1 = 22 + 273 = 295°K
T2 = 35 + 273 = 308°K
2.- Use the combine gas law to solve this problem
P1V1/T1 = P2V2/T2
-Solve for V2
V2 = P1V1T2 / T1P2
3.- Substitution
V2 = (1.5 x 2.5 x 308) / (295 x 3)
4.- Simplification
V2 = 1155 / 885
5.- Result
V2 = 1.3 l
C - nucleus, cytoplasm, plasma membrane.
The potential energy from the batteries in the torch is transferred to the filament of the bulb. The energy is transferred to the surroundings as thermal energy and light
Answer:
188.5g of dextrose are needed
Explanation:
In Weight per volume percentage - %(w/v) -, the concentration is defined as the mass of solute in grams -In this case, dextrose-, in 100mL of solution.
As you want to prepare 725mL of a 26.0% (w/v) solution. you need:
725mL * (26g / 100mL) = 188.5g of solute =
<h3>188.5g of dextrose are needed</h3>
