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1 year ago

4NH3+5O2=4NO+6H20 if 32.5 grams of NH3 react with enough oxygen, how many grams of water should form?

Chemistry

1 answer:

Aneli [31]1 year ago

6 0

51.3 gram of water should form if 32.5 g of NH₃ react with enough oxygen.

<h3>How to find the Number of moles ?</h3>

To calculate the number of moles use the formula

Number of moles = $\frac{\text{Given Mass}}{\text{Molar Mass}}$

= $\frac{32.5}{17}$

= 1.9 mol

4NH₃ + 5O₂ → 4NO + 6H₂O

4 mol of NH₃ react with oxygen to given 6 mole of water.

So 1.9 mol of NH₃ produces = $\frac{1.9}{4}\times 6$

= 2.85 mol of water

Mass of water = Molar Mass of water × Number of moles of water

= 18 × 2.85

= 51.3 gram

Thus from the above conclusion we can say that 51.3 gram of water should form if 32.5 g of NH₃ react with enough oxygen.

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4.65 L of nitrogen at standard pressure is compressed into a 0.480 L container. What is the new pressure in kPa?

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Assuming that nitrogen gas is ideal, we can use the equation PV = nRT to relate first conditions to the second condition. At constant temperature, pressure and volume are indirectly related as follows:

P = k / V

k is equal nRT

P1V1 = P2V2
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3 years ago

A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec

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<u>Answer:</u> The moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

<u>For lead (II) nitrate:</u>

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol$

<u>For NaI:</u>

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol$

For the given chemical reaction:

$Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, $3.14\times 10^{-5}$ moles of NaI will react with = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol$ of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

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

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, $3.14\times 10^{-5}$ moles of NaI will produce = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles$ of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

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Answer:

Land resources such as gravel and bedrock are used to construct buildings, roads, and sidewalks. Land is also where city structures—such as buildings, roads, and sidewalks—are constructed. People use land resources such as bedrock and aggregate to construct buildings, roads, and sidewalks.

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Answer: On losing 6 moles of water, cobalt chloride forms unstable violet-coloured ions, before generating its stable blue-coloured anhydrous form.

Explanation:

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<h3>What is the speed of light?</h3>

Speed of light is the rate of speed though the light travels. To find the speed of light in any medium, the following formula is used.

$v=\dfrac{c}{n}$

Here, (n) is the index of reaction and (c) is the speed of light in the vacuum. The speed of light in the vacuum is almost equal to the 3.0×10⁸ m/s.

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$v=\dfrac{c}{n}\\vn=c\\c=vn$

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Thus, the quantity which is equivalent to the product of the absolute index of refraction of water and the speed of light in water system is the speed of light in vacuum.

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