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Marat540 [252]
4 years ago
15

The balanced equation for water is 2 H2 + O2 to 2 H2O. If I have 21.2g of a product , and I started with 5.6 g of H2, how many g

rams of oxygen were reacted
Chemistry
1 answer:
ANTONII [103]4 years ago
7 0

Since 21.2 g H2O was produced, the amount of oxygen that reacted can be obtained using stoichiometry. The balanced equation was given:  2H₂ + O₂ → 2H₂O and the molar masses of the relevant species are also listed below. Thus, the following equation is used to determine the amount of oxygen consumed.

Molar mass of H2O = 18 g/mol

Molar mass of O2 = 32 g/mol

21.2 g H20 x 1 mol H2O/ 18 g H2O x 1 mol O2/ 2 mol H2O x 32 g O2/ 1 mol O2 = 18.8444 g O2

<span>We then determine that 18.84 g of O2 reacted to form 21.2 g H2O based on stoichiometry. It is important to note that we do not need to consider the amount of H2 since we can derive the amount of O2 from the product. Additionally, the amount of H2 is in excess in the reaction.</span>

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Can someone please help with chemistry? Lead has a density of 10.5 g/cm^3. What is the diameter of a lead ball that has a mass o
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9. How long will it take 120 grams of Strontium-90 to decay to 7.5 grams?<br>with steps pls​
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Answer:

116 years

Explanation:

To solve this, we will use the half life equation;

A(t) = A_o(½)^(t/t_½)

Where;

A(t) is the amount of strontium left after t years;

A_o is the initial quantity of strontium that will undergo decay;

t_½ is the half-life of strontium

t is the time it will take to decay

We are given;

A(t) = 7.5 g

A_o = 120 g

From online values, half life of strontium-90 is 29 years. Thus, t_½ = 29

Thus;

7.5 = 120 × ½^(t/29)

Divide both sides by 120 to get;

7.5/120 = ½^(t/29)

0.0625 = ½^(t/29)

In 0.0625 = (t/29) In ½

-2.772589 = (t/29) × (-0.693147)

(t/29) = -2.772589/(-0.693147)

t/29 = 4

t = 29 × 4

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