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To form 100 g of #H_2O_2#, 5.927 g of H must react with 94.073 g of O. How many grams of O would be needed to form 104 g of #H_2O_2#?

Answer 1

Well, what is #5.927%# of #104*g#?

And I make this....

#5.927%xx104*g=6.164*g#

And let us just check this...and examine the empirical formula of the #104*g# mass...and for this we divide thru by the ATOMIC mass of each constituent element...

#"Moles of hydrogen"=(6.164*g)/(1.00794*g*mol^-1)=6.116*mol#

#"Moles of oxygen"=(104*g-6.164*g)/(16.00*g*mol^-1)=6.116*mol#

And thus the empirical formula is simply #HO#...why is this different from the chemical formula of hydrogen peroxide?

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Answer 2

Caleb Adams

5.927 g of H and 94.073 g of O react to form 100 g of H2O2, so the ratio of H to O is 5.927:94.073, or roughly 1:15.89. If we were to calculate the amount of O needed for 104 g of H2O2, it would be 104 g / 100 g * 94.073 g = 97.76 g.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.