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Whitepunk [10]
3 years ago
8

How many moles of oxygen are required to produce 37.15 g CO2?

Chemistry
2 answers:
qwelly [4]3 years ago
8 0

Answer : The number of moles of oxygen required are, 0.844 moles

Solution : Given,

Mass of carbon dioxide = 37.1 g

Molar mass of carbon dioxide = 44 g/mole

First we have to calculate the moles of carbon dioxide.

\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}=\frac{37.15g}{44g/mole}=0.844moles

Now we have to calculate the moles of oxygen.

The balanced chemical reaction will be,

C+O_2\rightarrow CO_2

From the balanced reaction we conclude that

As, 1 mole of carbon dioxide produces 1 mole of oxygen

So, 0.844 mole of carbon dioxide produces 0.844 mole of oxygen

Therefore, the number of moles of oxygen required are, 0.844 moles

gayaneshka [121]3 years ago
3 0
Carbon = 12.010
Oxygen = 15.999 x 2
15.999 x 2 = 31.998 + 12.010 = 44.008
\frac{37.15 grams * 1 mole CO2}{44.008 grams}
Cancel out the grams
\frac{37.15 mol CO2}{44.008} = 0.844 mol CO2
2 moles of Oxygen atoms in 1 mol CO2 so
0.844 * 2 =
1.688 moles of Oxygen

You might be interested in
How many molecules of water are there in 0.234 L of water
tankabanditka [31]

Answer:

7.82x10^24 molecules of water

Explanation:

H2O=18.015 g/mol   Avogadro's #=6.022x10^23 molecules

0.234L x 1000g/1L x 1 mol H2O/18.015 g x 6.022x10^23 = 7.82x10^23 molecules of water

3 0
3 years ago
25 g of NH, is mixed with 4 moles of O, in the given reaction:
FromTheMoon [43]

Answer:

a. NH3 is limiting reactant.

b. 44g of NO

c. 40g of H2O

Explanation:

Based on the reaction:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l)

4 moles of ammonia reacts with 5 moles of oxygen to produces 4 moles of NO and 6 moles of water.

To find limiting reactant we need to find the moles of each reactant and using the balanced equation find which reactant will be ended first. Then, with limiting reactant we can find the moles of each reactant and its mass:

<em>a. </em><em>Moles NH3 -Molar mass. 17.031g/mol-</em>

25g NH3*(1mol/17.031g) = 1.47moles NH3

Moles O2 = 4 moles

For a complete reaction of 4 moles of O2 are required:

4mol O2 * (4mol NH3 / 5mol O2) = 3.2 moles of NH3.

As there are just 1.47 moles, NH3 is limiting reactant

b. Moles NO:

1.47moles NH3 * (4mol NO/4mol NH3) = 1.47mol NO

Mass NO -Molar mass: 30.01g/mol-

1.47mol NO * (30.01g/mol) = 44g of NO

c. Moles H2O:

1.47moles NH3 * (6mol H2O/4mol NH3) = 2.205mol H2O

Mass H2O -Molar mass: 18.01g/mol-

2.205mol H2O * (18.01g/mol) = 40g of H2O

3 0
3 years ago
Read 2 more answers
Analysis of a compound of sulfur, oxygen and fluorine showed that it is 31.42% S and 31.35% O, with F accounting for the remaind
Leya [2.2K]

Answer:

The molecular formula is  SO2F2

Explanation:

Step 1: Data given

Suppose the mass of compound = 100 grams

The compound contains:

31.42 % S = 31.42 grams S

31.35 % O = 31.35 grams O

100 - 31.42 - 31.35 = 37.23 F

Molar mass of S = 32.065 g/mol

Molar mass F = 19.00 g/mol

Molar mass O = 16.00 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles S = 31.42 grams / 32.065 g/mol

Moles S = 0.9799 moles

Moles 0 = 31.35 grams / 16.00 g/mol

Moles 0 = 1.959 moles

Moles F = 37.23 grams / 19.00 g/mol

Moles F = 1.959 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

S: 0.9799 / 0.9799 = 1

F: 1.959/ 0.9799 = 2

O : 1.959 / 0.9799 = 2

The empirical formula is SO2F2

This formula has a molecular mass of 102.06 g/mol

This means the empirical formula is also the molecular formula : SO2F2

3 0
3 years ago
Suppose 215 g of NO3- flows into a swamp each day. What volume of CO2 would be produced each day at 17.0°C and 1.00 atm?
charle [14.2K]

Answer:

The answer is "41.23 \ L\  N_2"

Explanation:

2 NO_3^{-} + 10 e^{-} + 12 H^{+} \longrightarrow N_2 + 6 H_2O\\\\= \frac{( 215 \ g \ NO_3^{-})}{(62.0049  \frac{\ g NO_3^{-}}{mol})} \times  \frac{(1 \ mol \ N_2}{ 2 \ mol \ NO_3^{-})}\\\\

=3.46746789 \times 0.5\\\\= 1.733 \ mol \ N_2 \\\\\to V = \frac{nRT}{P} \\\\= (1.733 \ mol) \times (0.08205746 \frac{L\ atm}{Kmol}) \times \frac{ (17 + 273) K}{(1.00 atm)}\\\\= 41.23

8 0
3 years ago
How many grams of iron can be made with 21.5g of Fe2O3
SIZIF [17.4K]

The mass (in grams) of iron, Fe that can be made from 21.5 g of Fe₂O₃ is 15.04 g

We'll begin by writing the balanced equation for the reaction. This is given below:

2Fe₂O₃ -> 4Fe + 3O₂

  • Molar mass of Fe₂O₃ = 159.7 g/mol
  • Mass of Fe₂O₃ from the balanced equation = 2 × 159.7 = 319.4 g
  • Molar mass of Fe = 55.85 g/mol
  • Mass of Fe from the balanced equation = 4 × 55.85 = 223.4 g

From the balanced equation above,

319.4 g of Fe₂O₃ decomposed to produce 223.4 g of Fe

<h3>How to determine the mass of iron, Fe produced</h3>

From the balanced equation above,

319.4 g of Fe₂O₃ decomposed to produce 223.4 g of Fe

Therefore,

21.5 g of Fe₂O₃ will decompose to produce = (21.5 × 223.4) / 319.4 = 15.04 g of Fe

Thus, 15.04 g of Fe were produced.

Learn more about stoichiometry:

brainly.com/question/9526265

#SPJ1

5 0
1 year ago
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