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SCORPION-xisa [38]
3 years ago
14

In the reaction below, how many grams of h2o(g) are produced when 2.1 grams o2(g) are consumed? c4h6(g) + o2(g) → co2(g) + h2o(g

)
Chemistry
2 answers:
Sergio [31]3 years ago
4 0
<span>0.54 g
       
Lookup the atomic weights of oxygen and hydrogen. Atomic weight oxygen = 15.999 Atomic weight hydrogen = 1.00794 Calculate molar masses Molar mass O2 = 2 * 15.999 = 31.998 g/mol Molar mass H2O = 2 * 1.00794 + 15.999 = 18.01488 g/mol Now determine how many moles of O2 you have 2.1 g / 31.998 g/mol = 0.065629102 mol Balance the equation so you get as the balanced equation, 2 C4H6(g) + 11 O2(g) → 8 CO2(g) + 6 H2O(g) Looking at the balanced equation, for every 11 moles of O2 consumed, 5 moles of H2O are produced. So divide the number of moles of O2 you have by 11, then multiply by 5, giving: 0.065629102 mol / 11 * 5 = 0.02983141 mol And finally, multiply by the molar mass of H2O, so 0.02983141 mol * 18.01488 g/mol = 0.53740927 g Round the results to 2 significant figures getting 0.54 g</span>
amm18123 years ago
3 0
The balanced equation that illustrates the reaction is:
2C4H6 + 11O2 ......> 8CO2 + 6H2O 

number of moles = mass / molar mass 
number of moles of oxygen = 2.1 / 32 = 0.065625 moles

Now, from the balanced equation, we can note that:
11 moles of oxygen are required to produce 6 moles of water.
Therefore:
0.065625 moles of oxygen will produce:
(0.065625*6) / 11 = 0.03579 moles of water

number of moles = mass / molar mass
mass = number of moles * molar mass
mass of water = 0.03579 * 18 = 0.644 grams
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E = hf

E = hc/lamda

where E-energy,

h-planck's constant (6.62 x 10^-34 Js)

f- frequency ( f = c/lamda)

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lamda- wavelength

you can this formula to solve all the 4 questions.

8 0
3 years ago
Water can be formed in the following reaction:
Komok [63]

The  moles  of  oxygen  gas (O2) that is needed is    4  moles


 Explanation

2H2 +O2 → 2H2O

The  moles of O2  is determined using the mole  ratio  of H2:O2

that is from equation  above H2:O2  is 2:1  

If the moles of H2  is 8  moles therefore  the moles  of O2

 = 8 moles x 1/2  = 4 moles

3 0
3 years ago
What volume of carbon dioxide gas can be collected from
alisha [4.7K]

Answer:

1.22 L of carbon dioxide gas

Explanation:

The reaction that takes place is:

  • CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O

First we <u>determine which reactant is limiting</u>:

  • Calcium carbonate ⇒ 10.0 g CaCO₃ ÷ 100 g/mol = 0.10 mol CaCO₃
  • Hydrochloric acid ⇒ 0.100 L * 0.50 M = 0.05 mol HCl

So HCl is the limiting reactant.

Now we calculate the moles of CO₂ produced:

  • 0.05 mol HCl * \frac{1molCO_{2}}{1molHCl} = 0.05 mol CO₂

Finally we use PV=nRT to <u>calculate the volume</u>:

  • P = 1 atm
  • n = 0.05 mol
  • T = 25 °C ⇒ 25 + 273.16 = 298.16 K

1 atm * V = 0.05 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K

  • V = 1.22 L
7 0
3 years ago
Which element has the lowest first ionization energy? (F), (Li), (Fr), (Rb)?
Mariulka [41]

Answer:

Fr

Explanation:

Francium (Fr) has the lowest ionization energy, with a value of 4.0727eV. This ionization energy increases as one advances in a period, being greater in the group of nonmetals.

6 0
3 years ago
The expression of the theoretical yield (TY) in function of limiting reagent (LR) of a reaction is as follows: TY = ideal mole r
spin [16.1K]

<u>Answer:</u> The theoretical yield of acetanilide is 6.5 grams.

<u>Explanation:</u>

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

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

  • <u>For aniline:</u>

Given mass of aniline = 4.50\times 10^0=4.50g      (We know that:  10^0=1 )

Molar mass of aniline = 93.13 g/mol

Putting values in equation 1, we get:

\text{Moles of aniline}=\frac{4.50g}{93.13g/mol}=0.048mol

  • <u>For acetic anhydride:</u>

To calculate the mass of acetic anhydride, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Volume of acetic anhydride = (1.25\times \text{Mass of aniline})=1.25\times 4.50=5.625mL

Density of acetic anhydride = 1.08 g/mL

Putting values in above equation:

1.08g/mL=\frac{\text{Mass of acetic anhydride}}{5.625mL}\\\\\text{Mass of acetic anhydride}=(1.08g/mL\times 5.625mL)=6.08g

Given mass of acetic anhydride = 6.08 g

Molar mass of acetic anhydride = 102.1 g/mol

Putting values in equation 1, we get:

\text{Moles of acetic anhydride}=\frac{6.08g}{102.1g/mol}=0.06mol

The chemical equation for the reaction of aniline and acetic anhydride follows:

C_6H_5NH_2+CH_3COOCOCH_3\rightarrow C_6H_5NHCOCH_3+CH_3COOH

By Stoichiometry of the reaction:

1 mole of aniline reacts with 1 mole of acetic anhydride

So, 0.048 moles of aniline will react with = \frac{1}{1}\times 0.048=0.048mol of acetic anhydride

As, given amount of acetic anhydride is more than the required amount. So, it is considered as an excess reagent.

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

By Stoichiometry of the reaction:

1 mole of aniline produces 1 mole of acetanilide

So, 0.048 moles of aniline will produce = \frac{1}{1}\times 0.048=0.048mol of acetanilide

Now, calculating the theoretical yield of acetanilide by using equation 1:

Moles of acetanilide = 0.048 moles

Molar mass of acetanilide = 135.17 g/mol

Putting values in equation 1, we get:

0.048mol=\frac{\text{Mass of acetanilide}}{135.17g/mol}\\\\\text{Mass of acetanilide}=(0.048mol\times 135.17g/mol)=6.5g

Hence, the theoretical yield of acetanilide is 6.5 grams.

3 0
4 years ago
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