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<u>Answer:</u> The limiting reagent is oxygen gas.
<u>Explanation:</u>
Limiting reagent is defined as the reactant that is present in less amount and it limits the formation of products.
Excess reagent is defined as the reactant which is present in large amount.
To calculate the number of moles, we use the equation:
.....(1)
- <u>For propane:</u>
Given mass of propane = 30.0 g
Molar mass of propane = 44.1 g/mol
Putting values in equation 1, we get:
- <u>For oxygen:</u>
Given mass of oxygen = 75.0 g
Molar mass of oxygen = 32 g/mol
Putting values in equation 1, we get:
The chemical equation for the combustion of propane follows:
By Stoichiometry of the reaction:
5 moles of oxygen gas reacts with 1 mole of propane.
So, 2.34 moles of oxygen gas will react with = of propane
As, given amount of propane is more than the required amount. So, it is considered as an excess reagent.
Thus, oxygen is considered as a limiting reagent because it limits the formation of product.
Hence, the limiting reagent is oxygen gas.
Answer:
215.12 g
Explanation:
The change in the bolling temperature is a collateral property called ebullioscopy. The elevation on temperature occurs because there'll be a solute in the pure compound, and that solute(generally non-volatile) has a different boiling point, so the total bolling pint should increase.
To calculate the elevation in the temperature, we must use the Raoult equation:
ΔT = KbxWxi
Where <em>ΔT</em> is the variation of the temperature of boiling point, <em>Kb</em> is the
ebullioscopy constant, <em>W</em> is the molarity, and <em>i</em> the Vant' Hoff factor, which for molecules is also equal to 1, because they don't form ions. So:
82.00 - 78.40 = 1.22xW
W = 3.60/1.22
W = 2.9508
The molarity can also be calculated by:
Where <em>m1</em> is the mass of the solute (in grames), <em>M1</em> is the molar mass of the solute, and <em>m2</em> is the mass of the solvent (in kg).
The density of ethanol is 0.789 g/cm³, and 1 mL = 1 cm³, so 600.00 mL = 600.00 cm³ :
0.789 = m2/600
m2 = 473.4 g = 0.4734 kg
The molar mass of geraniol is:
C: 12 g/mol x 10 = 120 g/mol
H: 1 g/mol x 18 = 18 g/mol
O: 16 g/mol x 1 = 16 g/mol
M1 = 154 g/mol
Then:
m1 = 215.12 g of geraniol.