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BaLLatris [955]
3 years ago
11

If 15 g of C₂H₆ reacts with 60.0 g of O₂, how many moles of water (H₂O) will be produced?

Chemistry
1 answer:
IceJOKER [234]3 years ago
5 0

Answer:

n_{H_2O}=1.5molH_2O

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O

Next, we identify the limiting reactant by computing the available moles of ethane and the moles of ethane consumed by 60.0 grams of oxygen:

n_{C_2H_6}^{available}=15g*\frac{1mol}{30g} =0.50molC_2H_6\\n_{C_2H_6}^{reacted}=60.0gO_2*\frac{1molO_2}{32gO_2}*\frac{2molC_2H_6}{7molO_2} =0.536molC_2H_6

Thus, we notice there are less available moles, for that reason, the ethane is the limiting reactant. Finally, we can compute the produced moles of water by:

n_{H_2O}=0.50molC_2H_6*\frac{6molH_2O}{2molC_2H_6}\\\\n_{H_2O}=1.5molH_2O

Best regards.

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Answer:

1.99 x 10⁻¹⁸J

Explanation:

Given parameters:

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Unknown:

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To solve this problem, we use the expression below;

     E  = hf

Where E is the energy, h is the Planck's constant and f is the frequency

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6 0
3 years ago
the pressure of a sample of argon gas was increased from 2.32 atm2.32 atm to 7.16 atm7.16 atm at constant temperature. if the fi
SOVA2 [1]

initial volume of the argon sample = 5.93L according to Boyle's law

What is Boyle's law ?

Boyle's law, also known as Mariotte's law, is a relationship describing how a gas will compress and expand at a constant temperature. The pressure (p) of a given quantity of gas changes inversely with its volume (v) at constant temperature, according to this empirical connection, which was established by the physicist Robert Boyle in 1662. In equation form, this means that pv = k, a constant.

According to Boyle's law

P1/V1 = P2/V2

P1 = initial pressure

P2 = final pressure

V1 =initial volume

V2= final volume

V1 = P1*V2/P2

V1 = 2.32*18.3/7.16 = 5.93L

initial volume of the argon sample = 5.93L according to Boyle's law

To know about Boyle's law from the link

brainly.com/question/26040104

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6 0
1 year ago
What is the limiting reagent when a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen?​
UNO [17]

Answer:

Ammonia is limiting reactant

Amount of oxygen left  = 0.035 mol

Explanation:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Balance chemical equation:

4NH₃ + 3O₂     →     2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

                      NH₃          :            N₂

                        4             :             2

                      0.12           :           2/4×0.12 = 0.06

                      NH₃         :            H₂O

                        4            :             6

                        0.12       :           6/4×0.12 = 0.18

                       

                       O₂            :            N₂

                        3             :             2

                      0.125        :           2/3×0.125 = 0.08

                        O₂           :            H₂O

                        3              :             6

                        0.125       :           6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

                        NH₃          :             O₂

                           4            :              3

                           0.12       :          3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left  = 0.125 - 0.09 = 0.035 mol

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