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Hitman42 [59]
3 years ago
5

From the following balanced equation, CH4(g)+2O2(g)⟶CO2(g)+2H2O(g) how many grams of H2O can be formed when 1.25g CH4 are combin

ed with 1.25×10^23 molecules O2? Use 6.022×10^23 mol−1 for Avogadro's number.
Chemistry
1 answer:
Montano1993 [528]3 years ago
7 0

Answer:

2.81 g of H2O.

Explanation:

We'll begin by calculating mass of O2 that contains 1.25×10²³ molecules O2.

This can be obtained as follow:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.022×10²³ molecules. This implies that 1 mole of O2 also contains 6.022×10²³ molecules.

1 mole of O2 = 16x2 = 32 g.

Thus 6.022×10²³ molecules is present in 32 g of O2,

Therefore, 1.25×10²³ molecules will be present in =

(1.25×10²³ × 32) / 6.022×10²³ = 6.64 g of O2.

Therefore, 1.25×10²³ molecules present in 6.64 g of O2.

Next, the balanced equation for the reaction. This is given below:

CH4(g) + 2O2(g) —> CO2(g) + 2H2O(g)

Next, we shall determine the masses of CH4 and O2 that reacted and the mass of H2O produced from the balanced equation.

This can be obtained as follow:

Molar mass of CH4 = 12 + (4x1) = 16 g/mol.

Mass of CH4 from the balanced equation = 1 x 16 = 16 g

Molar mass of O2 = 16x2 = 32 g/mol.

Mass of O2 from the balanced equation = 2 x 32 = 64 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol.

Mass of H2O from the balanced equation = 2 x 18 = 36 g

From the balanced equation above,

16 g of CH4 reacted with 64 g of O2 to produce 36 g if H2O.

Next, we shall determine the limiting reactant.

This can be obtained as follow:

From the balanced equation above,

16 g of CH4 reacted with 64 g of O2.

Therefore, 1.25 g of CH4 will react with = (1.25 x 64)/16 = 5 g of O2.

From the above calculations, we can see that only 5 g out of 6.64 g of O2 is needed to react completely with 1.25 g of CH4.

Therefore, CH4 is the limiting reactant.

Finally, we shall determine the mass of H2O produced from the reaction.

In this case, the limiting reactant will be used because it will give the maximum yield of H2O.

The limiting reactant is CH4 and the mass of H2O produced from the reaction can be obtained as follow:

From the balanced equation above,

16 g of CH4 reacted to produce produce 36 g if H2O.

Therefore, 1.25 g of CH4 will react to produce = (1.25 x 36)/16 = 2.81 g of H2O.

Therefore, 2.81 g of H2O were obtained from the reaction.

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Give the structure of the principal organic product formed by free-radical bromination of 2,2,4−trimethylpentane.
Fudgin [204]

the principal organic product formed by free-radical bromination of 2,2,4−trimethylpentane is 2-bromo-2,4,4-trimethylpentane.

what is free radical halogenation?

A substitution reaction in which a hydrogen atom is replaced with a halogen atom, via a free radical reaction mechanism. when this reaction is carrid out by bromine radical, it is called free radicle bromination. When bromine (Br{2}) treated with light (hν) it comes to homolytic cleavage of the Br-Br bond and give rise to bromine radicles.

free-radical bromination of 2,2,4−trimethylpentane

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Br - Br ⇒ 2Br·

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the principal organic product formed by free-radical bromination of 2,2,4−trimethylpentane is 2-bromo-2,4,4-trimethylpentane.

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gizmo_the_mogwai [7]

Answer:

Explanation:

1) ZnBr₂ (aq) + AgNO₃ (aq)

Chemical equation:

 ZnBr₂ (aq) + AgNO₃ (aq)  →Zn(NO₃)₂(aq) + AgBr(s)

Balanced chemical equation:

ZnBr₂ (aq) + 2AgNO₃ (aq)  →Zn(NO₃)₂(aq) + 2AgBr(s)

Ionic equation:

Zn²⁺(aq) + Br₂²⁻ (aq) + 2Ag⁺ (aq)+ 2NO⁻₃ (aq)  → Zn²⁺(aq) +(NO₃)₂²⁻(aq) + 2AgBr(s)

Net ionic equation:

Br₂²⁻ (aq) + 2Ag⁺ (aq)   →    2AgBr(s)

The Zn²⁺((aq) and NO⁻₃ (aq) are spectator ions that's why these are not written in net ionic equation. The AgBr can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

2) HgCl₂ (aq) + KI (aq)  →

Chemical equation:

HgCl₂ (aq) + KI (aq)  → KCl + HgI₂

Balanced chemical equation:

HgCl₂ (aq) + 2KI (aq)  → 2KCl(aq) + HgI₂(s)

Ionic equation:

Hg²⁺(aq)  + Cl₂²⁻  (aq) + 2K⁺(aq) + 2I⁻ (aq)  →  HgI₂ (s) + 2K⁺(aq) + 2Cl⁻ (aq)

Net ionic equation:

Hg²⁺(aq)  + 2I⁻ (aq) →   HgI₂ (s)

The Cl⁻ ((aq)  and K⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The HgI₂ (s) can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

 

3) Ca(OH)₂ (aq) + Na₂SO₄ (aq)

Chemical equation:

Ca(OH)₂ (aq) + Na₂SO₄ (aq)  →   CaSO₄(s) + NaOH(aq)

Balanced chemical equation:

Ca(OH)₂ (aq) + Na₂SO₄ (aq)  →   CaSO₄(s) + 2NaOH(aq)

Ionic equation:

Ca²⁺(aq)  + OH₂²⁻  (aq) + 2Na⁺(aq) + SO₄²⁻ (aq)  →   CaSO₄(s) + 2Na⁺(aq) + 2OH⁻ (aq)

Net ionic equation:

Ca²⁺(aq)   + SO₄²⁻ (aq)  →   CaSO₄(s)

The OH⁻ ((aq)  and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The CaSO₄ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

4 0
3 years ago
POSSIB
siniylev [52]

Answer:

96.99 C degree change

Explanation:

  480 cal  / ( 150 g * .033 cal/g-C )  = 96.99 C

8 0
2 years ago
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