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2 years ago

No links please. How many moles of methane (CH4) are in 7.31x10^25 molecules?

Chemistry

1 answer:

ad-work [718]2 years ago

4 0

Answer:

molar mass of methane CH4

= C + 4 H

= 12.0 + 4 x 1.008

= 12.0 + 4.032

= 16.042g/mol

7.31 x 10^25 molecules x 1 mole CH4 = 121.43 moles

6.02 x 10^23 CH4 molecules

121.43 moles CH4 are present.

Explanation:

not to certain if this is right or not.. but hope it helps!

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Upon combustion, a 0.8376 g sample of a compound containing only carbon, hydrogen, and oxygen produces 1.6003 gco2 and 0.9827 gh

Wewaii [24]

Answer is: empirical formula for compound is C₂H₆O.
m(CO₂) = 1,6003 g.
n(CO₂) = m(CO₂) ÷ M(CO₂).
n(CO₂) = 1,6003 g ÷ 44 g/mol.
n(CO₂) = 0,0364 mol.
n(CO₂) = n(C).
m(C) = 0,0364 mol · 12 g/mol = 0,4368 g.
m(H₂O) = 0,9827 g.
n(H₂O) = 0,9827 g ÷ 18 g/mol.
n(H₂O) = 0,0546 mol.
n(H) = 2 · n(H₂O) = 0,1092 mol.
m(H) = 0,1092 mol · 1 g/mol = 0,1092 g.
m(O) = 0,8376 g - 0,4368 g - 0,1092 g = 0,2916 g.
n(O) = 0,2916 g ÷ 16 g/mol = 0,0182 mol.
n(C) : n(H) : n(O) = 0,0364 mol : 0,1092 mol : 0,0182 mol.
n(C) : n(H) : n(O) = 2 : 6 : 1.

7 0

4 years ago

When iron(III) oxide reacts with hydrochloric acid, iron(III) chloride and water are formed. How many grams of iron(III) chlorid

Aleksandr [31]

Answer: The mass of iron (III) chloride produced is 14.81 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For iron(III) oxide:

Given mass of iron(III) oxide = 10.0 g

Molar mass of iron(III) oxide = 159.7 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron(III) oxide}=\frac{10.0g}{159.7g/mol}=0.0626mol$

For hydrochloric acid:

Given mass of hydrochloric acid = 10.0 g

Molar mass of hydrochloric acid = 36.5 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrochloric acid}=\frac{10.0g}{36.5g/mol}=0.274mol$

The chemical equation for the reaction of iron (III) oxide and hydrochloric acid follows:

$Fe_2O_3+6HCl\rightarrow 2FeCl_3+3H_2O$

By Stoichiometry of the reaction:

6 moles of hydrochloric acid reacts with 1 mole of iron (III) oxide

So, 0.274 moles of hydrochloric acid will react with = $\frac{1}{6}\times 0.274=0.0456mol$ of iron (III) oxide

As, given amount of iron (III) oxide is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

6 moles of hydrochloric acid produces 2 moles of iron (III) chloride

So, 0.274 moles of hydrochloric acid will produce = $\frac{2}{6}\times 0.274=0.0913moles$ of iron (III) chloride

Now, calculating the mass of iron (III) chloride from equation 1, we get:

Molar mass of iron (III) chloride = 162.2 g/mol

Moles of iron (III) chloride = 0.0913 moles

Putting values in equation 1, we get:

$0.0913mol=\frac{\text{Mass of iron (III) chloride}}{162.2g/mol}\\\\\text{Mass of iron (III) chloride}=(0.0913mol\times 162.2g/mol)=14.81g$

Hence, the mass of iron (III) chloride produced is 14.81 grams

7 0

3 years ago

Hydrofluoric acid solutions cannot be stored in glass containers because HF reacts readily with silica dioxide in glass to produ

satela [25.4K]

A) Limiting reactant

You need the molar ratios (from the balanced chemical equation) and the molar masses of each compound (from the atomic masses)

a) Molar ratios:

6 mol HF : 1 mol SiO2 : 1 mol H2SiF6

2) Molar masses:

Atomic masses:
H: 1 g/mol
F: 19 g/mol
Si: 28 g/mol
O: 16g/mol

=>
HF:1g/mol + 19 g/mol = 20 g/mol
SiO2: 28g/mol + 2*16g/mol = 60 g/mol
H2SiF6: 2*1g/mol + 28g/mol + 6*19g/mol = 144g/mol

3) convert data in grams to moles

21.0 g SiO2 / 60 g/mol = 0.35 mol SiO2

70.5 g HF / 20 g/mol = 3.525 mol HF

4) Use the theorical ratios to deduce which is in excess and which is the limiting reactant.

6 mol HF / 1mol SiO2 < 3.525 mol HF / 0.35 mol SiO2 ≈ 10

=> There is more HF than the needed to react with 0.35mol of SiO2 =>

SiO2 is the limiting reactant (HF is in excess)

b) Mass of excess reactant.

1) Calculate how many grams reacted, which requires to calculate first the number of moles that reacted

0.35 mol SiO2 * 6 mol HF / 1 mol SiO2 = 2.1 mol of HF

2.1 mol HF * 20 g/mol = 42 gram of HF

2) Subtract the quantity that reacted from the original quantity:

70.5 g - 42 g = 28.5 g of HF in excess

c) Theoretical yield of H2SiF6

1 mol of SiO2 ; 1 mol of H2SiF6 => 0.35 mol SiO2 : 0.35 mol H2SiF6

Convert those moles to grams: 0.35 mol * 144 g/mol = 50.4 grams

d) % yield

% yield = actual yield / theoretical yield * 100 = 45.8 / 50.4 * 100 = 90.87%

4 0

4 years ago

Consider the following oxidation-reduction reaction: 2fe3+(aq) + 2hg(l) + 2cl−(aq) → 2fe2+(aq) + hg2cl2(s). indicate the oxidizi

nirvana33 [79]

Answer:

The reducing agent is Hg (mercury) and the oxidixing agent is Cl (chlorine)

Explanation:

first find the oxidation charges of each of the different ions. All elements that are alone have a charge of 0. As you compare the reactants and products we can see that the chargers of Hg and Cl have changed.

Hg went from a charge of 0 to a charge of +1

Cl went from a charge of 0 to a charge of -1

remember when a substance gains electrons it is reducing and when a substance is losing electrons it is oxidizing.

use the acryonm

OIL

Oxidizing Is Losing

RIG

Reducing Is Gaining

8 0

2 years ago

How does solid neon compare to liquid neon?

Lapatulllka [165]

Answer:

The liquid molecules stay towards the bottom but one or two do make it all the way to the top. The molecules in the gas are also moving faster than the molecules in the liquid phase.

7 0

3 years ago

Read 2 more answers