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vladimir1956 [14]
2 years ago
7

Given the following equation: 2h20=2h2+02 how many grams of o2 are produced if 3.6 grams of h20 react?

Chemistry
1 answer:
inn [45]2 years ago
3 0

Answer:

3.2 g O₂

Explanation:

To find the mass of O₂, you need to (1) convert grams H₂O to moles H₂O (via molar mass), then (2) convert moles H₂O to moles O₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles O₂ to grams O₂ (via molar mass). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 2 sig figs to reflect the sig figs of the given value (3.6 g).

Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol

Molar Mass (H₂O): 18.014 g/mol

2 H₂O -----> 2 H₂ + 1 O₂

Molar Mass (O₂): 2(15.998 g/mol)

Molar Mass (O₂): 31.996 g/mol

3.6 g H₂O         1 mole               1 mole O₂           31.996 g
----------------  x  ---------------  x  ---------------------  x  ---------------  = 3.2 g O₂
                         18.014 g          2 moles H₂O          1 mole

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Darya [45]

Answer:

See Explanation

Explanation:

The equation of the reaction;

KHSO4(aq) + KOH(aq) -------> K2SO4(aq) + H2O(l)

Number of moles of KHSO4 = 49.6 g/136.169 g/mol = 0.36 moles

Since the reaction is in a mole ratio of 1:1, 0.36 moles of K2SO4 is produced.

Number of moles of KOH = 25.3 g/56.1056 g/mol = 0.45 moles

Since the reaction is 1:1, 0.45 moles of K2SO4 is produced

Hence K2SO4 is the limiting reactant.

Mass of K2SO4 formed = 0.36 moles of K2SO4 * 174.26 g/mol = 62.7 g

So;

1 mole of KHSO4 reacts with 1 mole of KOH

0.36 moles of KHSO4 reacts with 0.36 * 1/1 = 0.36 moles of KOH

Amount of excess KOH = 0.45 moles - 0.36 moles = 0.09 moles

Mass of excess KOH = 0.09 moles * 56.1056 g/mol  = 5 g of excess KOH

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3 years ago
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Explanation:

Given:

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