Question

Question #e4858

Answer 1

Balanced equation:
`C_3` `H_8`(g) + 5`O_2`(g) ___> 3 C`O_2`(g) + 4`H_2`O(g)

as per the equation , one mole of `C_3` `H_8` produces 4 moles of `H_2`O .

I can write this as ( 1 mol `C_3` `H_8` / 4 moles of`H_2`O )

I have started with 5 g of Propane. In terms of moles # of moles of propane = mass of propane / molar mass of propane

moles = (5g / 44 g `mol^(-1)`)

moles of propane = 0.114 moles.

0.114 moles of `C_3` `H_8` will produce , X moles of `H_2`O

I can write this as ( 0.114 mol `C_3` `H_8` / X moles of `H_2`O)

The two ratios are equal ( for making a cake I need two eggs and 3 cups of sugar.For making three such cakes , I will need 6 eggs and 9 cups of sugar, the ratio of sugar to flour will remain the same , it will not change with the # of cakes I want to make, but what will definitely change is the amount of the ingredients, but the ratio will stay the same.) Using the same logic the two ratios are equal.

( 1 mol `C_3` `H_8` / 4 moles of`H_2`O ) = ( 0.114 mol `C_3` `H_8` / X moles of `H_2`O)

Cross multiply
1 . x = 4 x 0.114
x = 0.456moles
0.114 moles ( or 5g ) of `C_3` `H_8`will react / need 0.456 moles of `H_2`O to completely react as per above chemical reaction.

one mole of water has mass 18 g / `mol^(-1)` so, 0.456 mole will weigh 0.456 mole x18 g / `mol^(1)` = 8.208 g. This was the expected yield. In this experiment only 80 % of what expected was obtained. So actual yield is 80 % of the expected tield
so the amount of water actually obtained was 0.80 x 8.208 = 6.556 g.