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Arisa [49]
3 years ago
6

Caluclate the number of grams of oxygen that must react

Chemistry
1 answer:
bonufazy [111]3 years ago
8 0

Answer : The number of grams of oxygen react must be, 170.4 grams.

Solution : Given,

Mass of C_3H_8 = 46.85 g

Molar mass of C_3H_8 = 44 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_3H_8.

\text{ Moles of }C_3H_8=\frac{\text{ Mass of }C_3H_8}{\text{ Molar mass of }C_3H_8}=\frac{46.85g}{44g/mole}=1.065moles

Now we have to calculate the moles of O_2

The balanced chemical reaction is,

C_3H_8+5O_2\rightarrow 3CO_2+4H_2O

From the balanced reaction we conclude that

As, 1 mole of C_3H_8 react with 5 mole of O_2

So, 1.065 moles of C_3H_8 react with 1.065\times 5=5.325 moles of O_2

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(5.325moles)\times (32g/mole)=170.4g

Therefore, the number of grams of oxygen react must be, 170.4 grams.

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Explanation:

8 0
3 years ago
How many moles are in 1.50 grams of ethanol, CH3CH2OH?
LekaFEV [45]
CH3CH2OH = 46.068 (molar mass)
1.50g(1 mole / 46.0680) =
0.03256 moles
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3 years ago
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The vapor pressure of water at 25.0°C is 23.8 torr. Determine the mass of glucose (molar mass = 180 g/mol) needed to add to 500.
mrs_skeptik [129]

According to Raoult's law the relative lowering of vapour pressure of a solution made by dissolving non volatile solute is equal to the mole fraction of the non volatile solute dissolved.

the relative lowering of vapour pressure is the ratio of lowering of vapour pressure and vapour pressure of pure solvent

\frac{p^{0-}p}{p^{0}}=x_{B}

Where

xB = mole fraction of solute=?

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x_{B}=\frac{23.8-22.8}{23.8}=0.042

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putting the values

molefraction=\frac{molessolute}{molesolute+molessolvent}

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Im sure the answer is c

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