Sig figs chemistry help CHEMISTRY MATH IDK

What is the volume of ammonia produced at 243 K at a pressure of 1.38 atm by the unbalanced reaction on the left if 5740 moles o

Neporo4naja [7]

Answer:

49671 L is the produced volume of ammonia

Explanation:

We think the reaction of ammonia 's production:

N₂(g) + 3H₂(g) → 2NH₃ (g)

We have the moles of each reactant so let's determine the limiting reactant:

Ratio is 1:3. 1 mol of nitrogen reacts with 3 moles of H₂

Then, 1720 moles of N₂ will react with (1720 .3) /1 = 5160 moles of H₂

We have 5740 moles of hydrogen, so we have enough hydrogen. This is the excess reagent, so the limiting is the N₂

1 mol of N₂ can produce 2 moles of ammonia

Therefore 1720 moles of N₂ will produce (1720 . 2) /1 = 3440 moles of NH₃

We apply now, the Ideal Gases Law → P . V = n . R .T

V = (n . R . T) /P → V = (3440 mol . 0.082 L.atm/mol.K . 243K) / 1.38 atm

V = 49671 L

We confirm that the nitrogen was the limiting reactant

3 moles of H₂ need 1 mol of nitrogen to react

Therefore, 5740 moles of H₂ will react with (5740 . 1) /3 = 1913 moles of N₂

It was ok to say, that N₂ was the limiting reactant because we need 1913 moles in the reaction, and we only have 1720 moles

6 0

2 years ago

Hospital patients are administered oxygen from an pressurized

VashaNatasha [74]

Answer:

32000atm

Explanation:

Using Boyle's law equation;

P1V1 = P2V2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (

V2 = final volume (L)

According to the question below:

P1 = 160.0 atm

P2 = 3.0 atm

V1 = 600L

V2 = ?

Using P1V1 = P2V2

160 × 600 = 3 × V2

96000 = 3V2

V2 = 96000/3

V2 = 32000atm

5 0

2 years ago

2c4H10 +13O2 —>8CO2+10H2O to find how many moles of oxygen would react with 4.3 mol C4H10

Elena-2011 [213]

Answer:

$\large\boxed{\text{28 mol of O$_{2}$}}$

Explanation:

2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

n/mol: 4.3

13 mol of O₂ react with 2 mol of 2C₄H₁₀

$\text{Moles of O}_{2} = \text{4.3 mol C$_{4}$H$_{10}$} \times \dfrac{\text{13 mol O}_{2}}{\text{2 mol C$_{4}$H$_{10}$}} = \textbf{28 mol O}_{2}\\\\\text{The reaction requires $\large\boxed{\textbf{28 mol of O$_{2}$}}$}$

5 0

2 years ago

As a sample's temperature increases, which two factors also increase? A. Particle speed B. Particle kinetic energy C. Particle b

jeyben [28]

Answer:

C.

Explanation:

It’s particle boiling point because the atoms are moving fast around the particle as possible, so there for its C..

8 0

2 years ago

Read 2 more answers

If 10.00 g of iron metal is burned in the presence of excess of O2 how many grams of Fe2O3 will form

sergiy2304 [10]

14.292 grams of Fe2O3 is formed when 10 gram of iron metal is burned.

Explanation:

The balanced equation for the reaction is to be known so that number of moles taking part can be known.

The balanced chemical equation is

4Fe + 3 $O_{2}$ ⇒ 2 $Fe{2}$ $O{3}$

From the given weight of iron to be used for the production of $Fe{2}$ $O{3}$ , number of moles of Fe taking part in the reaction can be known by the formula:

Number of moles= mass ÷ Atomic mass of one mole of the element.

(Atomic weight of Fe is 55.845 gm/mole)

Putting the values in equation

Number of moles = 10 gm ÷ 55.845 gm/mole

= 0.179 moles

Applying the stoichiometry concept

4 moles of Fe gives 2 Moles of Fe2O3

0.179 moles will produce x moles of Fe2O3

So, 2÷ 4 = x ÷ 0.179

2/4 = x/ 0.179

2 × 0.179 = 4x

2 × 0.179 / 4 = x

x = 0.0895 moles

So from 10 grams of iron metal 0.0895 moles of Fe2O3 is formed.

Now the formula used above will give the weight of Fe2O3

weight = atomic weight × number of moles

= 159.69 grams × 0.0895

= 14.292 grams of Fe2O3 formed.

4 0

2 years ago