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Jobisdone [24]
1 year ago
7

When methane ( CH4 ) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reac

tion isCH4(g)+O2(g)→CO2(g)+H2O(g) This type of reaction is referred to as a complete combustion reaction.What mass of water is produced from the complete combustion of 5.90×10−3 g of methane?Express your answer with the appropriate units.
Chemistry
1 answer:
slega [8]1 year ago
5 0
<h2>Answer:</h2>1.33*10^{-2}grams

<h2>Explanations</h2>

The complete balanced equation for the given reaction is expressed as;

CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

Given the following parameters

Mass of CH4 = 5.90×10^−3 g = 0.0059grams

Determine the moles of methane

\begin{gathered} moles\text{ of CH}_4=\frac{mass}{molar\text{ mass}} \\ moles\text{ of CH}_4=\frac{0.0059}{16.04} \\ moles\text{ of CH}_4=0.000368moles \end{gathered}

According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:

\begin{gathered} moles\text{ of H}_2O=\frac{2}{1}\times0.000368 \\ moles\text{ of H}_2O=0.000736moles \end{gathered}

Determine the mass of water produced

\begin{gathered} Mass\text{ of H}_2O=moles\times molar\text{ mass} \\ Mass\text{ of H}_2O=0.000736\times18.02 \\ Mass\text{ of H}_2O=0.0133grams=1.33\times10^{-2}grams \end{gathered}

Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams

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To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 2.2-L bulb, then filled
Ber [7]

Answer:

N2

Explanation:

We use the ideal gas equation to calculate the number of moles of the diatomic gas. Then from the number of moles we can get

Given:

P = 2atm

1atm = 101,325pa

2atm = 202,650pa

T = 27 degrees Celsius = 27 + 273.15 = 300.15K

V = 2.2L

R = molar gas constant = 8314.46 L.Pa/molK

PV = nRT

Rearranging n = PV/RT

Substituting these values will yield:

n = (202,650 * 2.2)/(8314.46* 300.15)

n = 0.18 moles

To get the molar mass, we simply divide the mass by the number of moles.

5.1/0.18 = 28.5g/mol

This is the closest to the molar mass of diatomic nitrogen N2.

Hence, the gas is nitrogen gas

7 0
3 years ago
I need A through D, I have no idea on what I'm supposed to do
geniusboy [140]

a. 48.6 is magnesium and 32.0 is oxygen

b. 80.6

c. also 80.6

d. yes, because the product has equal mass to the reactants

3 0
3 years ago
An experiment shows that a 236 mL gas sample has a mass of 0.443 g at a pressure of 740 mmHg and a temperature of 22 ∘C. What is
aleksley [76]

Answer:

49.2 g/mol

Explanation:

Let's first take account of what we have and convert them into the correct units.

Volume= 236 mL x (\frac{1 L}{1000 mL}) = .236 L

Pressure= 740 mm Hg x (\frac{1 atm}{760 mm Hg})= 0.97 atm

Temperature= 22C + 273= 295 K

mass= 0.443 g

Molar mass is in grams per mole, or MM= \frac{mass}{moles} or MM= \frac{m}{n}. They're all the same.

We have mass (0.443 g) we just need moles. We can find moles with the ideal gas constant PV=nRT. We want to solve for n, so we'll rearrange it to be

n=\frac{PV}{RT}, where R (constant)= 0.082 L atm mol-1 K-1

Let's plug in what we know.

n=\frac{(0.97 atm)(0.236 L)}{(0.082)(295K)}

n= 0.009 mol

Let's look back at MM= \frac{m}{n} and plug in what we know.

MM= \frac{0.443 g}{0.009 mol}

MM= 49.2 g/mol

3 0
3 years ago
El oxígeno y el azufre reaccionan con el cobre para formar óxido de cobre (CuO) y sulfuro de cobre (CuS), respectivamente. ¿Qué
oksian1 [2.3K]
Yes it need to be like that cause when it like that it like that
5 0
3 years ago
If 842 grams of sodium hydroxide reacts with 750.0 grams of aluminum, how many grams of aluminum hydroxide should theoretically
Phantasy [73]

548.55 grams of aluminum hydroxide should theoretically form.

Explanation:

Balanced equation for the reaction:

3 NaOH + Al ⇒ Al(OH)3 +3 Na

DATA GIVEN:

mass of NaOH = 842 grams, atomic mass =39.9 grams/mole

mass of Al = 750 grams, atomic mass = 26.9 grams/mole

aluminum hydroxide theoretical yield = ?

Moles of NaOH reacted

number of moles = \frac{mass}{atomic mass of 1 mole}

putting the values in the equation

NaOH = \frac{842}{39.9}

           = 21.1 MOLES OF NaOH

Al = \frac{750}{26.9}

   = 27.8 moles

from the equation

 from 3 moles of NaOH 1 mole of Al(OH)3 is produced

21.1 moles of NaOH will react to give x moles of Al(OH)3

\frac{1}{3} = \frac{x}{21.1}

7.03 moles of Al(OH)3 is formed.

and

1 mole of Al(OH)3 is formed from 1 mole of Al in the reaction

so, 27.8 Moles will react to give give 27.8 moles of Al(OH)3 limiting reagent of the given reaction is NaOH

mass of Al(OH)3 =7.03 x 78 (atomic mass of Al(OH)3)

          = 548.55 grams

theoretical  yield from the given data is 548.55 grams

3 0
3 years ago
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