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Karolina [17]
3 years ago
11

Hydrogen cyanide (HCN) is an important industrial chemical. It is produced from methane (CH4), ammonia, and molecular oxygen. Th

e reaction also produces water. Balance the equation for this reaction.
Chemistry
1 answer:
monitta3 years ago
4 0
In setting up and balancing equations, it is important to know the chemical formula of the substances involved and also it is important that the number of elements in both sides should be equal. We do as follows:

2CH4 + 2NH3 + 3O2 = 2HCN + 6H2O

Hope this helps.
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Use the equation editor or "Insert Chemistry - WIRIS editor" to write the balanced molecular chemical equation for the reaction
11Alexandr11 [23.1K]

Answer:

<u>Balanced equation:</u>

Pb(NO_{3})_{2}(aq)+K_{2}CO_{3}(aq)\rightarrow PbCO_{3}(s)+2KNO_{3}(aq)

Explanation:

The chemical reaction between Lead(II) Nitrate and potassium carbonate is as follows.

Lead(II)Nitrate+Potassium\,carbonate \rightarrow Lead(III)\,\,carbonate+Potassium\,nitrate

Pb(NO_{3})_{2}(aq)+K_{2}CO_{3}(aq)\rightarrow PbCO_{3}(s)+2KNO_{3}(aq)

<u>Ionic equation:</u>

Pb^{2+}(aq)+2NO_{3}^{-}(aq)+2K^{+}(aq)+CO_{3}^{2-}(aq)\Leftrightarrow PbCO_{3}(s)+K^{+}(aq)+2NO_{3}^{-}

Cancel the same ions on the both sides of the reaction.

The net ionic equation is as follows.

Pb^{2+}(aq)+CO_{3}^{2-}(aq)\Leftrightarrow PbCO_{3}(s)

4 0
3 years ago
In one of his experiments, Lavoisier placed 10.0 grams of mercury (II) oxide into a sealed container and heated it. The mercury
inysia [295]

Oxygen gas produced : 0.7 g

<h3>Further explanation</h3>

Given

10.0 grams HgO

9.3 grams Hg

Required

Oxygen gas produced

Solution

Reaction⇒Decomposition

2HgO(s)⇒2Hg(l)+O₂(g)

Conservation of mass applies to a closed system, where the masses before and after the reaction are the same

mass of reactants = mass of products

mass  HgO = mass Hg + mass O₂

10 g = 9.3 g + mass O₂

mass O₂ = 0.7 g

4 0
3 years ago
If 20.0 grams of Al is placed into a solution containg
meriva

Answer:

m H2(g) = 2.241 g H2(g)

Explanation:

  • 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g)

limit reagent:

∴ Mw Al = 26.982 g/mol

∴ Mw H2SO4 = 98.0785 g/mol

⇒ n Al = (20 g Al)×(mol/26.982 g) = 0.7412 mol Al

⇒ n H2SO4 = ( 115 g H2SO4 )×(mol/98.0785 g) = 1.173 mol H2SO4

⇒ n H2 = (0.7412 mol Al)×(3 mol H2/ 2 mol Al) = 1.112 mol H2

∴ Mw H2 = 2.016 g/mol

⇒ g H2 = (1.112 mol H2)×(2.016 g/mol) = 2.241 g H2

5 0
3 years ago
1. Sodium hydroxide reacts with carbon dioxide according to the equation: 2NaOH(s) + CO2(g) →
Leno4ka [110]

The limiting reagent when 5 g of NaOH and 4.4 g CO₂  allowed to react will be NaOH

<h3>What is Limiting reagent ?</h3>

The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.

Given chemical equation in balanced form ;

2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l).

According to the Chemical equation ;

  • The limiting reagent when 5 g of NaOH and 4.4 g CO₂  allowed to react will be NaOH

    If 44 g CO₂ requires 80 g of NaOH, therefore, 4.4 g CO₂ will require atleast 8 g of NaOH.

    But the available quantity is 5 g NaOH. thus, NaOH is the Limiting reagent.

  • 6.625 g of Na₂CO₃ are expected to be produced 5.0 g of NaOH and 4.4 g of CO₂ are allowed to react

    As 80 g NaOH produces 106 g of Na₂CO₃.

    Therefore 5 g NaoH will produce ;

    106 / 80 x 5 = 6.625 g

Learn more about limiting reagent here ;

brainly.com/question/11848702

#SPJ1

5 0
2 years ago
1. What kinds of metals are REALLY reactive? why?
Olin [163]

Answer:

1. Alkali metals (group 1)

2. halogens (Group 17)

3. noble gasses (group 18)

Explanation:

1. alkali metals only have one valence electron meaning that they really want to lose that one valence electron to get a full octet.

2. halogens have 7 valence electrons meaning that they just need to gain 1 to get a full octet.

3. Nobel gasses already have a full octet meaning that they don't want to react. (atoms only react to get a full octet)

I hope this helps.  Let me know if anything is unclear.

8 0
3 years ago
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