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3 years ago

What is the limiting reactant when 20g CH4 react with 15g H20?

Chemistry

1 answer:

Art [367]3 years ago

4 0

Answer:

H₂O.

Explanation:

It is clear from the balanced equation:

CH₄ + 2H₂O → CO₂ + 4H₂.

that 1.0 mole of CH₄ reacts with 2.0 moles of H₂O to produce 1.0 mole of CO₂ and 4.0 moles of H₂.

To determine the limiting reactant, we should calculate the no. of moles of (20 g) CH₄ and (15 g) H₂O using the relation:

n = mass/molar mass

no. of moles of CH₄ = mass/molar mass = (20 g)/(16 g/mol) = 1.25 mol.

no. of moles of H₂O = mass/molar mass = (15 g)/(18 g/mol) = 0.833 mol.

from the balanced reaction, 1.0 mole of CH₄ reacts with 2.0 moles of H₂O.

So, from the calculated no. of moles: 0.4167 mole of CH₄ reacts completely with 0.833 mole of H₂O and the remaining of CH₄ will be in excess.

So, the limiting reactant is H₂O.

You might be interested in

The balanced equation for the generation of sugar from sunlight water and co2 is

asambeis [7]

The balanced equation for the generation of sugar from sunlight water and CO₂ is

6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂

carbon dioxide + water → sugar + oxygen

The process of photosynthesis occurs when the chlorophyll present in the leaves of plants absorb sunlight to make food in the presence of carbon dioxide (enters through the stomata of leaves) and water (absorbed from the roots). As a result of this reaction sugar and oxygen is formed. After that sugar is converted in to starch and oxygen is released into air.

8 0

4 years ago

What would the volume of a gas be at 150c if had of volume of 693 ml at 45 c

Vlad1618 [11]

Answer:

Explanation:

T1 = 150°C = (150 + 273.15)K = 423.15K

T2 = 45°C = (45 + 273.15)K = 318K

V1 = 693mL = 693cm³

Applying Charle's law, the volume of a given gas is directly proportional to is temperature provided that pressure remains constant.

V = kT

V1 / T1 = V2 / T2

693 / 423.15 = V2 / 318

V2 = (693 * 318) / 423.15 = 520.79cm³

The new volume of the gas is 520.79cm³

6 0

3 years ago

When 7.80 mL of 0.500 M AgNO3 is added to 6.25 mL of 0.300 M NH4Cl, how many grams of AgCl are formed?

irina1246 [14]

Answer:

The answer to your question is 0.269 grams of AgCl

Explanation:

Data

[AgNO₃] = 0.50 M

Vol AgNO₃ = 7.80 ml

[NH₄Cl] = 0.30 M

Vol NH₄Cl = 6.25 ml

mass of AgCL

Balanced reaction

AgNO₃(aq) + NH₄Cl(aq) ⇒ AgCl (s) + NH₄NO₃ (aq)

Process

1.- Calculate the moles of AgNO₃

Molarity = moles / volume

moles = Molarity x volume

moles = 0.50 x 0.0078

moles = 0.0039

2.- Calculate the moles of NH₄Cl

moles = 0.30 x 0.0063

moles = 0.00188

3.- Calculate the limiting reactant

The proportion of AgNO₃(aq) to NH₄Cl(aq) is 1 :1, then, we conclude that the limiting reactant is NH₄Cl(aq), because there are less amount of this reactant in the experiment.

4.- Calculate the moles of AgCl

1 mol of NH₄Cl ---------------- 1 mol of AgCl

0.00188 mol of NH₄Cl ------------- x

x = (0.00188 x 1) /1

x = 0.00188 moles of AgCl

5.- Calculate the grams of AgCl

molecular mass of AgCl = 108 + 35.5 = 143.5 g

143.5 grams of AgCl -------------- 1 mol

x -------------------------------------------0.00188 moles of AgCl

x = (0.00188 x 143.5) / 1

x = 0.269 grams of AgCl

8 0

3 years ago

An IV bag is labeled as 0.800% w/w sodium chloride in water solution with a density of 1.036g/mL. What is the concentration of s

Goryan [66]

Answer:

[NaCl] = 0.14 M

[NaCl] = 0.14 m

Mole fraction of NaCl → 2.48×10⁻³

Explanation:

This is a problem of concentration.

0.8% w/w means that in 100 g of solution, we have 0.8 g of solute, in this case NaCl.

Let's determine the volume with the density.

Solution density = Solution mass / Solution volume

1.036 g/mL = 100 g / Solution volume

Solution volume = 100 g / 1.036 g/mL → 96.5 mL

Let's calculate molarity (mol/L)

We convert the mass to moles (mass / molar mass)

0.8 g / 58.45 g/mol = 0.0137 moles

We must convert the volume of solution to L.

Molarity is mol/L, moles of solute in L of solution.

96.5 mL . 1L/1000 mL = 0.0965 L

0.0137 mol / 0.0965 L = 0.14 M

Let's determine molality and mole fraction.

Molality are moles of solute in 1 kg of solvent (mol/kg)

Mass of solution = Mass of solute + Mass of solvent

100 g = 0.8 g + Mass of solvent

100 g - 0.8 g = Mass of solvent → 99.2 g

Then, we must convert the mass of solvent to kg

99.2 g . 1kg / 1000 g = 0.0992 kg

Molality: 0.0137 mol / 0.0992 kg → 0.14 m

Mole fraction → moles of solute / moles of solute + moles of solvent

Let's find out the moles of solvent ( mass / molar mass)

99.2g / 18 g/mol = 5.511 mol

Total moles = 5.511 + 0.0137 → 5.5247 moles

Mole fraction = 0.0137 / 5.5247 → 2.48×10⁻³

8 0

4 years ago

Cd(s)+2HCI(aq)---- H2(g)+CdCl2(aq). what volume in liters of 0.81m HCI solution would be needed to fully react with 32.71g Cd. a

BigorU [14]

Answer:

0.718L of 0.81M HCl are required

Explanation:

Based on the reaction:

Cd(s)+2HCI(aq) → H2(g)+CdCl2(aq)

1 mol of Cd reacts with 2 moles of HCl

To solve this question we must, as first, find the moles of Cd. With the moles of Cd we can find the moles of HCl needed to react completely with the Cd. With the moles and the molarity we can find the volume:

Moles Cd -Molar mass: 112.411g/mol-:

32.71g * (1mol / 112.411g) = 0.2910 moles Cd

Moles HCl:

0.2910 moles Cd * (2 moles HCl / 1mol Cd) =

0.5820 moles HCl

Volume:

0.5820 moles HCl * (1L / 0.81moles) =

<h3>0.718L of 0.81M HCl are required</h3>

6 0

3 years ago