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Aleksandr [31]
2 years ago
13

2Al + 6HCl → 2AlCl3 + 3H2 If 85.0 grams of HCl react, how many moles of H2 are produced?

Chemistry
1 answer:
Murrr4er [49]2 years ago
8 0

Answer:

1.17 mol

Explanation:

Step 1: Write the balanced equation

2 Al + 6 HCl → 2 AlCl₃ + 3 H₂

Step 2: Calculate the moles corresponding to 85.0 g of HCl

The molar mass of HCl is 36.46 g/mol.

85.0 g × 1 mol/36.46 g = 2.33 mol

Step 3: Calculate the number of moles of H₂ produced from 2.33 moles of HCl

The molar ratio of HCl to H₂ is 6:3.

2.33 mol HCl × 3 mol H₂/6 mol H₂ = 1.17 mol H₂

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Answer:

As solute concentration increases, vapor pressure decreases.  

Step-by-step explanation:

As solute concentration increases, the number of solute particles at the surface of the solution increases, so the number of <em>solvent </em>particles at the surface <em>decreases</em>.

Since there are fewer solvent particles available to evaporate from the surface, the vapour pressure decreases.

C. and D. are <em>wrong</em>. The vapour pressure depends <em>only</em> on the number of particles. It does not depend on the nature of the particles.

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Considering 2N2H4(g) + N2O4(g) -&gt; 3N2(g) +4H2O(g)
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107.8

Explanation:

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Carbon fixation occurs during the light reactions.
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How many liters of oxygen gas, at standard
Karo-lina-s [1.5K]

Answer:

Explanation:

  • For the balanced reaction:

<em>4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)​.</em>

It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.

  • Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:

no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.

  • Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:

<em><u>Using cross multiplication:</u></em>

4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.

0.64 mol of Fe is needed to react with → ??? mol of O₂.

∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.

  • Finally, we can get the volume of oxygen using the information:

<em>It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).</em>

<em></em>

<em><u>Using cross multiplication:</u></em>

1 mol of O₂ occupies → 22.4 L, at STP conditions.

0.48 mol of O₂ occupies → ??? L.

∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.

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3 years ago
What do chemists measure concentration with?
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How to calcutate concentration of solution.
there is 12gram of solute in a 36 gram solution
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