Question

Caluclate the number of grams of oxygen that must react

with46.85g of C3H 8 to produce only carbondioxide
and water.

Answer 1

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}=(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* 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* \text{ Molar mass of }O_2`

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

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