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lesya692 [45]
3 years ago
10

1) A 10g sample of H2(g) reacts with a 22g sample of O2(g) according to

Chemistry
1 answer:
solmaris [256]3 years ago
7 0

Answer:

H₂ is excess reactant and O₂ the limiting reactant

Explanation:

Based on the chemical reaction:

2H₂(g) + O₂(g) → 2H₂O

<em>2 moles of H₂ react per mole of O₂</em>

<em />

To find limiting reactant we need to convert the mass of each reactant to moles:

<em>Moles H₂ -Molar mass: 2.016g/mol-:</em>

10g H₂ * (1mol / 2.016g) = 4.96 moles

<em>Moles O₂ -Molar mass: 32g/mol-:</em>

22g O₂ * (1mol / 32g) = 0.69 moles

For a complete reaction of 0.69 moles of O₂ are needed:

0.69mol O₂ * (2mol H₂ / 1mol O₂) = 1.38 moles of H₂

As there are 4.96 moles,

<h3>H₂ is excess reactant and O₂ the limiting reactant</h3>
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Answer:

Metal

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Explanation:

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3 years ago
. What mass of ammonium chloride must be added to 250. mL of water to give a solution with pH
Eduardwww [97]

The mass of ammonium chloride that must be added is : ( A ) 4.7 g

<u>Given data :</u>

Volume of water ( V )  = 250 mL = 0.25 L

pH of solution = 4.85

Kb = 1.8 * 10⁻⁵

Kw = 10⁻¹⁴

Given that the dissolution of NH₄Cl gives NH₄⁺⁺ and Cl⁻ ions the equation is written as :

NH₄CI  +  H₂O  ⇄  NH₃ + H₃O⁺

where conc of H₃O⁺

[ H₃O⁺ ] = \sqrt{Ka.C}   and Ka = Kw / Kb

∴ Ka = 5.56 * 10⁻¹⁰

Next step : Determine the concentration of H₃O⁺  in the solution

pH = - log [ H₃O⁺ ] = 4.85

∴ [ H₃O⁺ ] in the solution = 1.14125 * 10⁻⁵

Next step : Determine the concentration of NH₄CI in the solution

C = [ H₃O⁺ ]² / Ka

  = ( 1.14125 * 10⁻⁵ )² /  5.56 * 10⁻¹⁰

  = 0.359 mol / L

Determine the number of moles of NH₄CI in the solution

n = C . V

  = 0.359 mol / L  * 0.25 L =  0.08979 mole

Final step : determine the mass of ammonium chloride that must be added to 250 mL

mass = n * molar mass

         = 0.08979 * 53.5 g/mol

         = 4.80 g  ≈ 4.7 grams

Therefore we can conclude that the mass of ammonium chloride that must be added is 4.7 g

Learn more about ammonium chloride : brainly.com/question/13050932

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