*WILL GIVE BRAINLIEST TO CORRECT ANSWER * Question in picture

How many grams of carbon dioxide are produced when 16.0

Kobotan [32]

Mass of CO₂ = 33 g

<h3>Further explanation </h3>

Complete combustion of Hydrocarbons with Oxygen will be obtained by CO₂ and H₂O compounds.

If O₂ is insufficient there will be incomplete combustion produced by CO and H and O

Reaction

CH₄ + 2O₂⇒CO₂ + 2H₂O

mol CH₄ :

$\tt =\dfrac{16}{16}=1$

mol O₂ :

$\tt =\dfrac{48}{32}=1.5$

A method that can be used to find limiting reactants is to divide the number of moles of known substances by their respective coefficients

$\tt CH_4:O_2=\dfrac{1}{1}:\dfrac{1.5}{2}=1:0.75$

Because O₂ ratio smaller then O₂ becomes the limiting reactants

So mol CO₂ from limiting reactants

mol CO₂ :

$\tt \dfrac{1}{2}\times 1.5=0.75$

mass CO₂ :

$\tt mass=mol\times MW=0.75\times 44=33~g$

7 0

3 years ago

Read 2 more answers

Convert 6.62 X 10^23 atoms to moles

olga2289 [7]

Answer:

1.099

Explanation:

Hope this helps

5 0

3 years ago

Read 2 more answers

What depends on public opinion and support? A. regulations at a local level B. enforcement of laws C. success of a policy D. rev

Nastasia [14]

i believe the answer is success of a policy. if this was correct please mark brainliest and lmk if you have any more questions x

3 0

4 years ago

How many grams of Br are in 335g of CaBr2

ASHA 777 [7]

The answer is 267.93 g

Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol

The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%

Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g

7 0

3 years ago

A compound is 7.74% hydrogen and 92.26% carbon by mass. At 100°C a 0.6883 g sample of the gas occupies 250 mL when the pressure

ycow [4]

Answer: The molecular formula for the compound is $C_6H_6$

Explanation:

We are given:

Percentage of C = 92.26 %

Percentage of H = 7.74 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 92.26 g

Mass of H = 7.74 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{92.26g}{12g/mole}=7.68moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{7.74g}{1g/mole}=7.74moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 7.68 moles.

For Carbon = $\frac{7.68}{7.68}=1$

For Hydrogen = $\frac{7.74}{7.68}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula for the given compound is $CH$

Calculating the molar mass of the compound:

To calculate the molecular mass, we use the equation given by ideal gas equation:

PV = nRT

Or,

$PV=\frac{m}{M}RT$

where,

P = pressure of the gas = 820 torr

V = Volume of gas = 250 mL = 0.250 L (Conversion factor: 1 L = 1000 mL )

m = mass of gas = 0.6883 g

M = Molar mass of gas = ?

R = Gas constant = $62.3637\text{ L. torr }mol^{-1}K^{-1}$

T = temperature of the gas = $100^oC=(100+273)K=373K$

Putting values in above equation, we get:

$820torr\times 0.250L=\frac{0.6883g}{M}\times 62.3637\text{ L torr }mol^{-1}K^{-1}\times 373K\\\\M=\frac{0.6883\times 62.3637\times 373}{820\times 0.250}=78.10g/mol$