Based on the information, what is the balanced chemical equation representing the burning of butane?

OptionA. 2C4H10 + 13O2 —> 8CO2 + 10H20

Explanation:

Butane burns is air (O2) according to the equation:

C4H10 + O2 —> CO2 + H20

Considering the equation, it is evident that it not balanced. Now let us balance the equation as shown below;

There are a total of 4 carbon atoms on the left and 1 carbon atom on the right. It can be balanced by putting 4 in front of CO2 as shown below:

C4H10 + O2 —> 4CO2 + H20

Next, there are 10 hydrogen atoms on the left and 2 hydrogen atoms on the right. Therefore to balance it, put 5 in front of H2O as shown below:

C4H10 + O2 —> 4CO2 + 5H20

Now, there are a total of 13 oxygen atoms on the right and 2 at the left. To balance it, put 13/2 in front of O2

as shown below

C4H10 + 13/2O2 —> 4CO2 + 5H20

Now we multiply through by 2 clear off the fraction and we obtained:

2C4H10 + 13O2 —> 8CO2 + 10H20

43.452

step-by-step explanation:

284 times .153 = 43.452

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"