Question

How many molecules of H₂O end O2 are
presont in 8. sg of H2O2
​

Answer 1

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Answer 2

Answer:

Firstly consider the balanced chemical reaction that relates the two molecules:

O2 + 2H2O = 2H2O2

This formula tells us that 2 moles of H2O and 1 mole of O2 are required to make 2moles of H2O2.

Since we know that we have 8.5 grams of H2O2, it is best we work in grams in order to figure out how many grams of H2O were required to make such an amount.

Considering the perfect formula above,

M(H202) = 34.016g/mol

converting to grams: 34.016g/mol x 2mol = 68.032g

M(H20) = 18.016g/mol

converting to grams: 18.016g/mol x 2mol = 36.032g,  

we can see that 36.032g (slightly more than half of 68) of H2O is required to make 68.032g of H2O2.

By using ratio’s

(68.032 g H2O2 : 36.032g H2O

8.5g H2O2 : x g H2O )

and solving for x, it can be found that 4.502g of H2O is required for 8.5 g of H2O2. The mass of H2O required is also slightly more than half that of H2O2 in this case.

Now that we have solved for the amount of H2O required in grams, we can convert to moles and follow this by the multiplication of Avogadro’s number to find the number of molecules:

n (H20) = 4.502/18.016 = 0.249882 mol

Number of molecules: 0.249882 x 6.022*10^23 = 1.50479x10^23 molecules.

Therefore, there are 1.50479x10^23 molecules of H20 in 8.5g of H2O2.