Option C is the correct set of the problem for mass of water produced by 3.2 moles of oxygen and an excess ethene.
<h3>
Reaction between oxygen and ethene</h3>
Ethene (C2H4) burns in the presence of oxygen (O2) to form carbon dioxide (CO2) and water (H2O) along with the evolution of heat and light.
C₂H₄ + 3O₂ ----- > 2CO₂ + 2H₂O
from the equation above;
3 moles of O₂ ---------> 2(18 g) of water
3.5 moles of O₂ ----------> x
![x = 3.2 \times [\frac{2 \ moles \ H_2O}{3 \ moles \ O_2} ] \times[ \frac{18.02 \ g \ H_2O}{1 \ mole \ H_2O} ]](https://tex.z-dn.net/?f=x%20%3D%203.2%20%5Ctimes%20%5B%5Cfrac%7B2%20%5C%20moles%20%5C%20H_2O%7D%7B3%20%5C%20moles%20%5C%20O_2%7D%20%20%5D%20%5Ctimes%5B%20%5Cfrac%7B18.02%20%5C%20g%20%5C%20H_2O%7D%7B1%20%5C%20mole%20%5C%20H_2O%7D%20%5D)
Thus, option C is the correct set of the problem for mass of water produced by 3.2 moles of oxygen and an excess ethene.
Learn more about reaction of ethene here: brainly.com/question/4282233
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Answer:
0.0303 Liters
Explanation:
Given:
Mass of the potassium hydrogen phosphate = 0.2352
Molarity of the HNO₃ Solution = 0.08892 M
Now,
From the reaction it can be observed that 1 mol of potassium hydrogen phosphate reacts with 2 mol of HNO₃
The number of moles of 0.2352 g of potassium hydrogen phosphate
= Mass / Molar mass
also,
Molar mass of potassium hydrogen phosphate
= 2 × (39.09) + 1 + 30.97 + 4 × 16 = 174.15 g / mol
Number of moles = 0.2352 / 174.15 = 0.00135 moles
thus,
The number of moles of HNO₃ required for 0.00135 moles
= 2 × 0.00135 mol of HNO₃
= 0.0027 mol of HNO₃
Now,
Molarity = Number of Moles / Volume
thus,
for 0.0027 mol of HNO₃, we have
0.08892 = 0.0027 / Volume
or
Volume = 0.0303 Liters
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Answer:
0.252 milimoles
Explanation:
To convert mass of a substance to moles it is necessary to use the molar mass of the substance.
The formula of morphine is C₁₇H₁₉NO₃, thus, its molar mass is:
C: 17*12.01g/mol = 204.17g/mol
H: 19*1.01g/mol = 19.19g/mol
N: 1*14g/mol = 14g/mol
O: 3*16g/mol = 48g/mol.
204.17 + 19.19 + 14 + 16 = <em>285.36g/mol</em>
Thus, moles of 71.891 mg = 0.071891g:
0.071891g × (1mol / 285.36g) = 2.5193x10⁻⁴ moles
As 1 mole = 1000 milimoles:
2.5193x10⁻⁴ moles = <em>0.252 milimoles</em>
