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lions [1.4K]
3 years ago
7

PLEASE HELP ME!!! ASAP

Chemistry
1 answer:
shtirl [24]3 years ago
5 0

Answer:

Theoretical yield of the reaction = 34 g

Excess reactant is hydrogen

Limiting reactant is nitrogen

Explanation:

Given there is 100 g of nitrogen and 100 g of hydrogen

Number of moles of nitrogen = 100 ÷ 28 = 3·57

Number of moles of hydrogen = 100 ÷ 2 = 50

Reaction between nitrogen and hydrogen yields ammonia according to the following chemical equation

N2 + 3H2 → 2NH3

From the above chemical equation for every mole of nitrogen that reacts, 3 moles of hydrogen will be required and 2 moles of ammonia will be formed

Now we have 3·57 moles of nitrogen and therefore we require 3 × 3·57 moles of hydrogen

⇒ We require 10·71 moles of hydrogen

But we have 50 moles of hydrogen

∴ Limiting reactant is nitrogen and excess reactant is hydrogen

From the balanced chemical equation the yield will be 2 × 3·57 moles of ammonia

Molecular weight of ammonia = 17 g

∴ Theoretical yield of the reaction = 2 × 3·57 × 17 = 121·38 g

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4 0
2 years ago
How many atoms are in 10.1 g Ne
LekaFEV [45]

Answer:

3.01 × 10²³ atoms Ne

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Tables
  • Moles

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

<em>Identify</em>

[Given] 10.1 g Ne

[Solve] atoms Ne

<u>Step 2: Identify Conversions</u>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

[PT] Molar Mass of Ne: 20.18 g/mol\

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 10.1 \ g \ Ne(\frac{1 \ mol \ Ne}{20.18 \ g \ Ne})(\frac{6.022 \cdot 10^{23} \ atoms \ Ne}{1 \ mol \ Ne})
  2. [DA] Divide/Multiply [Cancel out units]:                                                          \displaystyle 3.01398 \cdot 10^{23} \ atoms \ Ne

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

3.01398 × 10²³ atoms Ne ≈ 3.01 × 10²³ atoms Ne

4 0
3 years ago
How do solve for #13?What is the boiling point of a solution made by dissolving 1.0000 mole of sucrose in 1.0000 kg of water?
exis [7]

What is the boiling point of a solution made by dissolving 1.0000 mole of sucrose in 1.0000 kg of water?

The change in Boiling Point of water can be calculated using this formula:

ΔTb = i * Kb * m

Where i is the van't hoff factor (the number of particles or ions), the kb is a constant (boiling point elevation constant) and m is the molality of the solution.

The kb for water is always 0.515 °C/m. Kb = 0.515 °C/m

The value for i in this case is 1. Since sucrose is a covalent compound and it doesn't dissociate into ions. i = 1

The molal concentration of the solution can be found using this formula:

molality = moles of sucrose/kg of water

molality = 1.000 mol / 1.000 kg of water

molality = 1 m

Now that we know all the values, we can use the formula to find the change in the boiling point of water:

ΔTb = i * Kb * m

ΔTb = 1 * 0.515 °C/m * 1 m

ΔTb = 0.515 °C

Finally, we are asked for the boiling point of the solution, not the change. The boiling point of water at atmospheric pressure is 100.00 °C. If the boiling point rises 0.515 °C when we prepare the solution. The boiling point of the solution is:

Boiling point solution = Boiling point of water + ΔTb

Boiling point solution = 100.000 °C + 0.515 °C

Boiling point solution = 100.515 °C

Answer: The boiling point of the solution is 100.515 °C.

8 0
1 year ago
How many moles of Cu do you need to react with 12 moles of O2?<br> 2Cu(s) + O2(g) 2CuO(s)
zzz [600]

Answer:

24 mol Cu

General Formulas and Concepts:

<u>Chemistry</u>

  • Stoichiometry

Explanation:

<u>Step 1: Define</u>

RxN: 2Cu (s) + O₂ (g) → 2CuO (s)

Given: 12 moles O₂

<u>Step 2: Stoichiometry</u>

<u />12 \ mol \ O_2(\frac{2 \ mol \ Cu}{1 \ mol \ O_2} ) = 24 mol Cu

<u>Step 3: Check</u>

<em>We are given 2 sig figs.</em>

Our final answer is in 2 sig figs, so no need to round.

3 0
3 years ago
What is the volume (in liters) of a 5.98 gram sample of O2 at STP?
borishaifa [10]

Answer:

4.186 L

Explanation:

Using the pv=nrt equation and converting the grams of O2 into mols. After finding the number of mols by dividing 5.98 by 32 (2*the atomic weight of O) you plug that into the equation. So then you have 1*V=.186875*.08206*273 then you rearrange the equation to solve for v and get 4.186 L

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3 years ago
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