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Fed [463]
3 years ago
14

7.32 moles of hydrogen reacts with 48.97 grams of nitrogen, how many moles of ammonia is produced?

Chemistry
1 answer:
faust18 [17]3 years ago
7 0

3.5 moles of ammonia (NH₃) are produced

Explanation:

We have the following chemical reaction where hydrogen (H₂) reacts with nitrogen (N₂) to produce ammonia (NH₃):

3 H₂ + N₂ → 2 NH₃

number of moles = mass / molecular weight

number of moles of N₂ = 48.97 / 28 = 1.75 moles

We see from the chemical reaction that 1 mole of N₂ will react with 3 moles of H₂, so 1.75 moles of nitrogen will react with 3 × 1.75 = 5.25 moles of H₂. We have 7.32 moles of H₂, a quantity more of what is needed, so the limiting reactant is N₂.

Knowing this we devise the following reasoning:

if         1 mole of N₂ produces 2 moles of NH₃

then   1.75 moles of N₂ produces X moles of NH₃

X = (1.75 × 2) / 1 = 3.5 moles of NH₃

Learn more about:

limiting reactant

brainly.com/question/7144022

#learnwithBrainly

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The following represents a lab set up for heat transfer. The cup on the left started with boiling water at 100 degrees C and the cup on the right has water at 20 degrees C. There is an aluminum bar between the two cups allowing heat to transfer from one cup into the other. The set up will be left alone for 20 minutes and temperatures of each cup of water will be recorded every minute for 20 minutes.

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URGENT CHEMISTRY EXPERT!
vovangra [49]

Answer:

Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)

Explanation:

Part 1. Volume of reactant

(a) Balanced chemical equation.

\rm 2Na_{3}PO_{4} + 3CuCl_{2} \longrightarrow Cu_{3}(PO_{4})_{2} + 6NaCl

(b) Moles of CuCl₂

\text{Moles of CuCl}_{2} =\text{ 16.7 mL CuCl}_{2} \times \dfrac{\text{0.200 mmol CCl}_{2}}{\text{1 mL CuCl}_{2}} =  \text{3.340 mmol CuCl}_{2}

(c) Moles of Na₃PO₄

The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

\text{Moles of Na$_{3}$PO}_{4} =  \text{3.340 mmol CuCl}_{2} \times \dfrac{\text{2 mmol Na$_{3}$PO}_{4}}{\text{3 mmol CuCl}_{2}} =\text{2.227 mmol Na$_{3}$PO}_{4}

(d) Volume of Na₃PO₄

V = \text{2.227 mmol Na$_{3}$PO}_{4}\times \dfrac{\text{1 mL Na$_{3}$PO}_{4}}{\text{0.300 mmol Na$_{3}$PO}_{4}} = \text{7.42 mL Na$_{3}$PO}_{4} \\\\\text{The reaction requires $\large \boxed{\textbf{7.42 mL Na$_{3}$PO}_{4}}$}

Part 2. Net ionic equation

(a) Molecular equation

\rm 2Na_{3}PO_{4}(\text{aq}) + 3CuCl_{2}(\text{aq}) \longrightarrow Cu_{3}(PO_{4})_{2}(\text{s}) + 6NaCl(\text{aq})

(b) Ionic equation

You write molecular formulas for the solids, and you write the soluble ionic substances as ions.

According to the solubility rules, metal phosphates are insoluble.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)  

(c) Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

<u>6Na⁺(aq)</u> + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + <u>6Cl⁻(aq)</u> ⟶ Cu₃(PO₄)₂(s) + <u>6Na⁺(aq)</u> + <u>6Cl⁻(aq)</u>  

The net ionic equation is

3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)

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