Calculate the mass of water produced when 1.36 g of butane reacts with excess oxygen.

Mass of water produced= 1.8 g

Explanation:

Given data:

Mass of water produced = ?

Mass of butane = 1.36 g

Mass of oxygen = excess

Solution:

Chemical equation:

2C₄H₁₀ +13 O₂       →   8CO₂ + 10H₂O

Number of moles of butane:

Number of moles = mass/molar mass

Number of moles = 1.36 g/ 58.12 g/mol

Number of moles = 0.02 mol

Now we will compare the moles of butane with water.

                 C₄H₁₀         :        H₂O

                  2              :           10

                 0.02          :         10/2×0.02 = 0.1 mol

Mass of water produced:

Mass = molar mass × molar mass

Mass = 0.1 mol × 18 g/mol

Mass = 1.8 g

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.

answer: i dont know the answer but this might

explanation:

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