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2 years ago

Methane (CH4), ammonia (NH3), and oxygen (O2) can react to form hydrogen cyanide (HCN) and water according to this equation:

Chemistry

1 answer:

umka2103 [35]2 years ago

7 0

Answer:

-----> 2 CH₄ + 2 NH₃ + 3 O₂ -----> 2 HCN + 6 H₂O

-----> CH₄ = Whichever reactant produces the smaller amount of product is the limiting reagent. This is because the limiting reagent runs out before all of the other reactant is completely used. Because CH₄ produces the smaller amount of product, it is the limiting reagent.

-----> 13.5 grams HCN

Explanation:

Part 1:

An equation is balanced when there is an equal amount of each reactant on both sides of the reaction. These amounts can be modified by adding coefficients in front of the molecules.

The unbalanced equation:

CH₄ + NH₃ + O₂ -----> HCN + H₂O

Reactants: 1 carbon, 7 hydrogen, 1 nitrogen, 2 oxygen

Products: 1 carbon, 3 hydrogen, 1 nitrogen, 1 oxygen

The balanced equation:

2 CH₄ + 2 NH₃ + 3 O₂ -----> 2 HCN + 6 H₂O

Reactants: 2 carbon, 14 hydrogen, 2 nitrogen, 6 oxygen

Products: 2 carbon, 14 hydrogen, 2 nitrogen, 6 oxygen

Part 2:

You can determine the limiting reactant by converting each reactant mass (besides O₂) to a product mass. Whichever reactant produces the smaller amount of product is the limiting reagent. This is because the limiting reagent runs out before all of the other reactant is completely used.

Let's convert to HCN because Part 3 also wants to know how much HCN is produced. To find this amount, you need to (1) convert grams reactant to moles reactant (via their molar masses), then (2) convert moles reactant to moles HCN (via the mole-to-mole ratio from equation coefficients), and then (3) convert moles HCN to grams HCN (via its molar mass).

Molar Mass (CH₄): 12.011 g/mol + 4(1.008 g/mol)

Molar Mass (CH₄): 16.043 g/mol

Molar Mass (NH₃): 14.009 g/mol + 3(1.008 g/mol)

Molar Mass (NH₃): 17.033 g/mol

Molar Mass (HCN): 1.008 g/mol + 12.011 g/mol + 14.009 g/mol

Molar Mass (HCN): 27.028 g/mol

8 g CH₄ 1 mole 2 moles HCN 27.028 g
-------------- x ---------------- x ---------------------- x ---------------- = 13.5 g HCN
16.034 g 2 moles CH₄ 1 mole

10 g NH₃ 1 mole 2 moles HCN 27.028 g
--------------- x ---------------- x ----------------------- x ---------------- = 15.9 g HCN
17.033 g 2 moles NH₃ 1 mole

Because CH₄ produces the smaller amount of product, it is the limiting reagent. Therefore, the actual amount of HCN produced is 13.5 g.

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3 years ago

A student combines 364.6 g of HCl with 80 g of NaOH in 5 L of water. What additional volume of H2O must be added to this mixture

Jobisdone [24]

Answer:

75L of additional water to have a pH 1 solution

Explanation:

The reaction of HCl With NaOH is:

HCl + NaOH → H₂O + NaCl

By using molar mass of each reactant you can know how many moles will react, thus:

HCl: 364.6g HCl ₓ (1mol / 36.46g) = 10 moles HCl

NaOH: 80g NaOH ₓ (1mol / 40g) = 2 moles NaOH

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A solution with pH = 1 contains:

pH = -log [H⁺]

1 = -log [H⁺]

0.1M = [H⁺]

As molarity, M is the ratio between moles and liters and you want a solution 0.1M having 8 moles of H⁺ you require:

0.1M = 8 moles H⁺ / 80L

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3 years ago

How many moles of k3po4 can be formed when 4.4 moles of h3po4 react with 3.8 moles of koh? h3po4 + koh yields h2o + k3po4 be sur

schepotkina [342]

the balanced chemical equation for the reaction is as follows

H₃PO₄ + 3KOH ---> K₃PO₄ + 3H₂O

stoichiometry of H₃PO₄ to KOH is 1:3

first we have to find which the limiiting reactant is

as the amount of product formed depends on the amount of limiting reactant present

number of H₃PO₄ moles reacted - 4.4 mol

if H₃PO₄ is the limiting reactant

1 mol of H₃PO₄ reacts with 3 mol of KOH

then 4.4 mol of H₃PO₄ reacts with - 3 x 4.4 mol = 13.2 mol of KOH

but only 3.8 mol of KOH is present

therefore KOH is the limiting reactant

stoichiometry of KOH to K₃PO₄ is 3:1

number of KOH moles reacted - 3.8 mol

therefore number of K₃PO₄ formed = number of KOH moles reacted / 3

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R = 0.082 atm·L·mol⁻¹·K⁻¹

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Inputing the data:

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And solving for n:

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Finally we convert N₂ moles into NaN₃ moles, using the stoichiometric coefficients of the balanced reaction:

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