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sveta [45]
3 years ago
9

Determine the volume (liters) of 0.750 m naoh solution required to neutralize 1.50 l of 0.500 m h2so4. the neutralization reacti

on is: h2so4 (aq) + 2naoh (aq) → na2so4 (aq) + 2 h2o (l)
Chemistry
1 answer:
Elena L [17]3 years ago
8 0
The balanced equation for the neutralisation reaction is as follows
2NaOH + H₂SO₄ --> Na₂SO₄  + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of H₂SO₄ moles reacted - 0.500 mol/L x 1.50 L = 0.750 mol
1 mol of H₂SO₄ reacts with 2 mol of NaOH
therefore 0.750 mol of H₂SO₄ reacts with - 2 x 0.750 = 1.50 mol of NaOH
molarity of NaOH solution is 0.750 M
there are 0.750 mol in 1 L of solution
therefore 1.50 mol in - 1.50 mol / 0.750 mol/L = 2 L

2 L of NaOH is required for neutralisation 


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WINSTONCH [101]

Density is defined as the ratio of mass to the volume.

Density = \frac{Mass}{Volume}          (1)

Mass of water = 10 grams

Mass of acetone  = 10 grams

Density of water  = 1 \frac{g}{cm^{3}}

Density of acetone  = 0.7857 \frac{g}{cm^{3}}

Put the value of density of water and its mass in equation (1)

1 \frac{g}{cm^{3}} =  \frac{10 g}{volume}

Volume of water =  10 cm^{3}

Put the value of density of acetone and its mass in equation (1)

0.7857 \frac{g}{cm^{3}} =  \frac{10 g}{volume}

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3 years ago
Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N
Montano1993 [528]

The question is incomplete, here is the complete question:

Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows: 2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 135.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 178.0 kg of urea.

Determine the limiting reactant. (express your answer as a chemical formula)

<u>Answer:</u> The limiting reactant is ammonia (NH_3)

<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 ammonia:</u>

Given mass of ammonia = 135.9 kg = 135900 g    (Conversion factor:  1 kg = 1000 g)

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonia}=\frac{135900g}{17g/mol}=7994.12mol

  • <u>For carbon dioxide gas:</u>

Given mass of carbon dioxide gas = 211.4 kg = 211400 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon dioxide gas}=\frac{211400g}{44g/mol}=4804.54mol

The given chemical reaction follows:

2NH_3(aq.)+CO_2(aq,)\rightarrow CH_4N_2O(aq.)+H_2O(l)

By Stoichiometry of the reaction:

2 moles of ammonia reacts with 1 mole of carbon dioxide

So, 7994.12 moles of ammonia will react with = \frac{1}{2}\times 7994.12=3997.06mol of carbon dioxide

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

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

Hence, the limiting reactant is ammonia (NH_3)

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<span><u><em>Answer:</em></u>
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<u><em>Explanation:</em></u>
<u>We can classify elements/compounds based on their pH values into three types:</u>
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<u>neutral:</u> these are compounds having pH value equal to 7
<u>alkalies:</u> these are compounds having pH values higher than 7

This is shown in the attached image

We are given that the pH of the compound ammonia generated by the fish is above 7.
According to the above explanation, compound ammonia would be an alkaline compound.

Hope this helps :)</span>

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