Answer:
There, there are no leftover for C6H14 instead, an additional mass of 4g of C6H14 is needed to completely react with 38.4g of O2.
Explanation:
Step 1:
We'll begin by writing the balanced equation for the reaction. This is shown below:
2C6H14 + 19O2 —> 12CO2 + 14H2O
Step 2:
Let us calculate the masses of C6H14 and O2 that reacted from the balanced equation. This is illustrated below:
2C6H14 + 19O2 —> 12CO2 + 14H2O
Molar Mass of C6H14 = (12x6) + (14x1) = 72 + 14 = 86g/mol
Mass of C6H14 from the balanced equation = 2 x 86 = 172g
Molar Mass of O2 = 16x2 =32g/mol
Mass of O2 from the balanced equation = 19 x 32 = 608g
From the balanced equation above, 172g of C6H14 reacted with 608g of O2.
Step 3.
Now, let us determine the mass of C6H14 that will react with 38.4 g of oxygen. This is illustrated below:
From the balanced equation above, 172g of C6H14 reacted with 608g of O2.
Therefore, Xg of C6H14 will react with 38.4g of O2 i.e
Xg of C6H14 = (172 x 38.4) /608
Xg of C6H14 = 10.9g
From the calculations made above, we can see clearly that the mass of C6H14 is limited as the reaction requires 10.9g of C6H14 and only 6.9g was given. There, there are no leftover for C6H14 instead, an additional mass ( 10.9 - 6.9 = 4g) of 4g of C6H14 is needed to completely react with 38.4g of O2.
