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mixer [17]
3 years ago
5

22.4l of ammonia is reaxts with 1.406 mole of oxygen to produce NO and h2o .1.what volume of no is produced at ntp​

Chemistry
1 answer:
IceJOKER [234]3 years ago
4 0

Answer:

The volume of NO is 22.4L at STP

Explanation:

Based on the reaction:

2NH3 + 5/2O2 → 2NO + 3H2O

<em>2 moles of NH3 react with 5/2 moles of O2 to produce 2 moles of NO.</em>

<em />

To solve this question, we need to find the moles of each reactant in order to find the limiting reactant as follows:

<em>Moles NH3 -Molar mass: -17.01g/mol-</em>

Using PV = nRT

PV/RT = n

<em>Where P is pressure = 1atm at STP</em>

<em>V is volume = 22.4L</em>

<em>R is gas constant = 0.082atmL/molK</em>

<em>T is absolute temperature = 273.15K</em>

1atm*22.4L/0.082atmL/molK*273.15K = n

n = 1.00 moles of NH3

For a complete reaction of 1.00 moles of NH3 are needed:

1.00 moles NH3 * (5/2moles O2 / 2moles NH3) = 1.25 moles of O2

As there are 1.406 moles of O2, <em>the limiting reactant is NH3</em>

<em />

The moles of NO produced are the same than moles of NH3 because 2 moles of NH3 produce 2 moles of NO. The moles of NO are 1.00 moles

And as 1.00moles of gas are 22.4L at STP:

<h3>The volume of NO is 22.4L at STP</h3>

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Answer:

We need 8.11 grams of glucose for this solution

Explanation:

Step 1: Data given

Molarity of the glucose solution = 0.300 M

Total volume = 0.150 L

The molecular weight of glucose = 180.16 g/mol

Step 2: Calculate moles of glucose in the solution

Moles glucose = molarity solution * volume

Moles glucose = 0.300 M * 0.150 L

Moles glucose = 0.045 moles glucose

Step 3: Calculate mass of glucose

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MAss glucose = 0.045 moles * 180.16 g/mol

MAss glucose = 8.11 grams

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A balloon is floating around outside your window. The temperature outside is 21 ∘C , and the air pressure is 0.700 atm . Your ne
Andreyy89

Answer:

V=162L

Explanation:

Hello,

In this case, we can compute the required volume by using the ideal gas equation as shown below:

PV=nRT

Thus, solving for the volume and considering absolute temperature (in Kelvins), we obtain:

V=\frac{nRT}{P}= \frac{4.70mol*0.082\frac{atm*L}{mol*K}*(21+273)K}{0.700atm} \\\\V=162L

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What is the volume of a balloon of gas at 842 mm Hg and -23° C, if its volume is 915 mL at a pressure of 1,170 mm Hg and a tempe
garik1379 [7]
Answer:
             V₂  =  1070 mL or 1.07 L

Solution:

Data Given;
                  P₁  =  1170 mmHg

                  V₁  =  915 mL

                  T₁  =  24 °C  +  273 K  =  297 K

                  P₂  =  842 mmHg

                  V₂  =  ?

                  T₂  =  - 23 °C  +  273 K  =  250 K

According to Ideal gas equation,

                       P₁ V₁ / T₁  =  P₂ V₂ / T₂

Solving for V₂,

                       V₂  =  P₁ V₁ T₂ / P₂ T₁

Putting Values,

                       V₂  = (1170 mmHg × 915 mL × 250 K) ÷ (842 mmHg × 297 K)

                       V₂  =  1070 mL or 1.07 L
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Answer:

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this is because heat is causing permanent changes and can no longer be changed back to its original atate

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