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poizon [28]
2 years ago
14

If a mixture containing 0.50 moles of N2 and 1.8 moles of H2 is allowed to react according to the equation N2 3H2 2NH3 Group of

answer choices N2 is the limiting reactant. Both A and B are correct. H2 is the limiting reactant. 34.0 grams of NH3 can be produced.
Chemistry
1 answer:
Lelechka [254]2 years ago
7 0

Answer: hydrogen is the limiting reactant.

Explanation:

We have the equation \text{N}_{2}+3\text{H}_{2} \longrightarrow 2\text{NH}_{3}.

This means that for every mole of nitrogen consumed, 3 moles of hydrogen are consumed.

  • Considering the nitrogen, the reaction can occur 0.50 times.
  • Considering the hydrogen, the reaction can occur 1.8/3 = 0.6 times.

Therefore, <u>hydrogen</u> is the limiting reactant.

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Oxygen gas can be prepared by heating potassium chlorate according to the following equation: 2KClO3(s)2KCl(s) + 3O2(g) The prod
Ipatiy [6.2K]

Answer:

The number of moles of KClO₃ reacted was 0,15 mol

Explanation:

For the reaction:

2KClO₃(s) → 2KCl(s) + 2O₂(g)

The only gas product is O₂.

Total pressure is the sum of vapor pressure of water with O₂ gas formed. Thus, pressure of O₂ is:

749mmHg - 23,8mmHg = 725,2mmHg

Using gas law:

PV/RT = n

Where:

P is pressure (725,2mmHg ≡ <em>0,9542atm</em>)

V is volume (<em>5,76L</em>)

R is gas constant (<em>0,082 atmL/molK</em>)

And T is temperature (25°C ≡ <em>298,15K</em>)

Replacing, number of moles of O₂ are <em>0,2248 moles</em>

As 2 moles of KClO₃ react with 3 moles of O₂ the moles of KClO₃ that reacted was:

0,2248 mol O₂×\frac{2 mol KClO_{3}}{3 mol O_{2}} = <em>0,15 mol of KClO₃</em>

I hope it helps!

8 0
3 years ago
Fill in the coefficients to balance the equation for the chemical reaction that occurs:
Dominik [7]

Answer:

The equation is balanced

Explanation:

NaCl (aq) + AgNO3(aq) ––> AgCl (s) + NaNO3 (aq)

NaCl (aq) + AgNO3 (aq)

Na = 1 , Cl=1 , Ag = 1 , No3= 1

AgCl (s) + NaNO3 (aq)

Ag = 1 , Cl=1 , Na = 1 , No3= 1

8 0
2 years ago
Which of the following are true statements about equilibrium systems? For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(
Grace [21]

Answer:

The first, third and fourth statements are correct.

Explanation:

1) For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(s) + CO2(g) adding more CaCO3 will shift the equilibrium to the right.

⇒ Le Chatellier says As the CaCO3 concentration is increased, the system will attempt to undo that concentration change by shifting the balance to the right. <u>This statement is true.</u>

<u />

2) For the following reaction at equilibrium: CaCO3(s)⇌ CaO(s) + CO2(g) increasing the total pressure by adding Ar(g) will shift the equilibrium to the right.

⇒ Le chatellier says that if we increase the pressure, the equilibrium will shift to the side with the least number of particles.

Since the molar densities of CaO and CaCO3 are constant, they don't appear in the equilibrium expression. This is why only changes to the pressure (concentration) of CO2 affect the position of the equilibrium.

If the pressure in the container is increased by adding an inert or non-reacting gas, nothing happens to the amounts of CO2, CaO or CaCO3. The added gas won't affect the partial pressure of CO2. <u>This statement is false. </u>

3)For the following reaction at equilibrium: 2 H2(g) + O2(g) ⇌ 2 H2O(g) the equilibrium will shift to the left if the volume is doubled.

⇒ Le Chatellier says if we increase the pressure, the equilibrium will shift to the side with the most particles.

In this case we have 2 moles of H2 and 1 mole of O2 on the left side and 2 mole of H2O on the right side. This means on the left side are more particles. So the equilibrium will shift to the left, so <u>this statement is true.</u>

4) For the following reaction at equilibrium: H2(g) + F2(g) ⇌ 2HF(g) removing H2 will increase the amount of F2 present once equilibrium is reestablished. Increasing the temperature of an endothermic reaction shifts the equilibrium position to the right.

⇒ Le chatellier says if H2 will be removed (this means the left side will get less particles) so the equilibrium will shift to the left, to increase the amount of F2.

⇒Le chatelier says if we increase the temperature of an exotherm reaction , there will be less energy released. The equilibrium will shift to the side of the reactants (the left side).

If we increase the temperature of an endotherm reaction, the equilibrium will shift to the side of the products (the right side). <u>This statement is true.</u>

4 0
3 years ago
Which cell type is represented by number 1?
alexira [117]

where is the graph i need it to answer haha put the graph then i will answer
4 0
3 years ago
Sodium hydrogen carbonate and citric acid<br><br> Word equation
timofeeve [1]

Answer:

the right answer is:

Explanation:

How to Balance: NaHCO3 + HC2H3O2 = NaC2H3O2 + CO2 + H2O|

5 0
3 years ago
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