Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 35.5 g of butane is mixed with 33. g of oxygen. Calculate the minimum mass of butane that could be left over by the chemical reaction. Round your answer to significant digits.

19. grams is the minimum mass of butane that could be left over by the chemical reaction.

Explanation:

Moles of butane =

Moles of oxygen =

According to reaction, 13 mole of oxygen  reacts with 2 moles of butane ,then 1.8 moles of oxygen will react with :

of butane

Moles of butane left unreacted = 0.61 mol - 0.28 mol = 0.33 mol

Mass of 0.3301 moles of butane :

0.33 mol × 58 g/mol = 19. g

19. grams is the minimum mass of butane that could be left over by the chemical reaction.

"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"