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Sergio [31]
3 years ago
15

Calculate the number of grams of carbon dioxide produced from 6 moles of oxygen gas. List only the number of grams of carbon dio

xide. First, you will need to balance the equation above. Round to the nearest hundredth. And, as always, don't forget the units.
Chemistry
1 answer:
faust18 [17]3 years ago
3 0

Answer:

g CO2(g) = 264.06 g

Explanation:

  • C(s)  + O2(g) → CO2(g)

∴ moles O2(g) = 6 mol

∴ mm CO2 = 44.01 g/mol

⇒ mol CO2(g) = ( 6 mol O2(g) )×( mol CO2(g) / mol O2(g) ) = 6 mol CO2(g)

⇒ mass CO2(g) = ( 6 mol CO2(g) )×( 44.01 g/mol ) = 264.06 g CO2(g)

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How many hydrogen atoms are in 35.0 grams of hydrogen gas? How many hydrogen atoms are in 35.0 grams of hydrogen gas? 4.25 × 102
Ede4ka [16]

Answer: 2.12\times 10^{25} atoms of hydrogen are there in

35.0 grams of hydrogen gas.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{35.0g}{2g/mol}=17.5moles

1 mole of hydrogen (H_2) = 2\times 6.023\times 10^{23}=12.05\times 10^{23} atoms

17.5 mole of hydrogen (H_2) = \frac{12.05\times 10^{23}}{1}\times 17.5=2.12\times 10^{25} atoms

There are 2.12\times 10^{25} atoms of hydrogen are there in

35.0 grams of hydrogen gas.

8 0
3 years ago
Glucose (C6H12O6) can be fermented to yield ethanol (CH3CH2OH) and carbon dioxide (CO2). C6H12O6⟶2CH3CH2OH+2CO2 The molar mass o
jekas [21]

Answer:

The % yield is 74.45 %

Explanation:

<u>Step 1:</u> The balanced equation

C6H12O6⟶2CH3CH2OH+2CO2

<u>Step 2</u>: Data given

Molar mass glucose = 180.15 g/mol

Molar mass of ethanol = 46.08 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Mass of glucose = 61.5 grams

Mass of ethanol = 23.4 grams

<u>Step 3:</u> Calculate moles of glucose

Moles glucose = Mass glucose  / Molar mass of glucose

Moles glucose = 61.5 grams / 180.15 g/mol

Moles glucose = 0.341 moles

<u>Step 4:</u> Calculate moles of ethanol

1 mole of glucose consumed, produces 2 moles of ethanol and 2 moles of CO2

0.341 moles of glucose, will produce 2*0.341 = 0.682 moles of ethanol

<u>Step 5:</u> Calculate mass of ethanol

Mass ethanol = moles ethanol * Molar mass ethanol

Mass ethanol = 0.682 moles * 46.08 g/mol

Mass ethanol = 31.43 grams = theoretical mass

<u>Step 6:</u> Calculate % yield

% yield = actual mass / theoretical mass

% yield = (23.4 grams / 31.43 grams) * 100%

% yield = 74.45 %

The % yield is 74.45 %

7 0
3 years ago
A 0.43g samle of KHP required 24.11cm of NaOH for neutralization. Calculate the molarity of NaOH
expeople1 [14]

Answer:

0.083 M

Explanation:

We'll begin by calculating the number of mole in 0.43 g of KHP (potassium hydrogen phthalate, C₈H₅O₄K). This is can be obtained as follow:

Mass of C₈H₅O₄K = 0.43 g

Molar mass of C₈H₅O₄K = (8×12) + (5×1) + (16× 4) + 39

= 96 + 5 + 64 + 39 = 204 g/mol

Mole of C₈H₅O₄K =?

Mole = mass / molar mass

Mole of C₈H₅O₄K = 0.43 / 204

Mole of C₈H₅O₄K = 0.002 mole

Next, we shall determine the number of mole of NaOH required to react with 0.43 g (i.e 0.002 mole) of KHP. This can be obtained as follow:

C₈H₅O₄K + NaOH → C₈H₄O₄KNa + H₂O

From the balanced equation above,

1 mole of KHP reacted with 1 mole of NaOH.

Therefore, 0.002 mole of KHP will also react with 0.002 mole of NaOH.

Next, we shall convert 24.11 cm³ to L. This can be obtained as follow:

1000 cm³ = 1 L

Therefore,

24.11 cm³ = 24.11 cm³ × 1 L / 1000 cm³

24.11 cm³ = 0.02411 L

Finally, we shall determine the molarity of NaOH. This can be obtained as follow:

Mole of NaOH = 0.00 2 mole

Volume = 0.02411 L

Molarity of NaOH =?

Molarity = mole /Volume

Molarity of NaOH = 0.002 / 0.02411

Molarity of NaOH = 0.083 M

3 0
3 years ago
Realiza los cálculos para determinar la cantidad de KOH 90%, que se necesita para preparar 100 ml de solución 1N.
attashe74 [19]

Answer:

6.23 KOH 90% son necesarios

Explanation:

Una solución 1N de KOH requiere 1equivalente (En KOH, 1eq = 1mol) por cada litro de solución.

Para responder esta pregunta se requiere hallar los equivalentes = Moles de KOH para preparar 100mL = 0.100L de una solución 1N. Haciendo uso de la masa molar de KOH y del porcentaje de pureza del KOH se pueden calcular los gramos requeridos para preparar la solución así:

<em>Equivalentes KOH:</em>

0.100L * (1eq / L) = 0.100eq = 0.100moles

<em>Gramos KOH -Masa molar: 56.1056g/mol-:</em>

0.100moles * (56.1056g/mol) = 5.61 KOH se requieren

<em>KOH 90%:</em>

5.61g KOH * (100g KOH 90% / 90g KOH) =

<h3>6.23 KOH 90% son necesarios</h3>
8 0
3 years ago
What is the difference between Arrhenius acids-base and bronsted-lowry acids bases? Justify your answer.
Oxana [17]

Answer:

A Bronsted-Lowry acid like and Arrhenius acid is a compound that breaks down to give an H+ in solution. The only difference is that the solution does not have to be water. ... An Arrhenius base is a molecule that when dissolved in water will break down to yield an OH- or hydroxide in solution.

Explanation:

7 0
3 years ago
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