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drek231 [11]
3 years ago
13

Given the reaction 2NaOH + H2SO4 = Na2SO4 + 2H2O, what is the total number of grams of NaOH needed to react completely with 196

grams of H2SO4
Chemistry
1 answer:
ollegr [7]3 years ago
3 0

Answer:

160 g of NaOH

Explanation:

Reaction: 2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O

When a base and an acid react, the produce water and a salt. → Neutralization

In this reaction 2 moles of hydroxide need 1 mol of sulfuric acid to react, it is the same to say that 1 mol of sulfuric acid needs 2 moles of NaOH to react. We convert the mass of sulfuric to moles and we propose the rule of three:

196 g / 98 g/mol = 2 moles

1 mol of sulfuric acid needs 2 moles of NaOH to react

Then 2 moles of sulfuric acid will react with (2 . 2) /1 = 4 moles of NaOH

We convert the moles of base to grams → 4 mol . 40 g /1mol = 160 g

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2.202 moles of ions are released when samples dissolves completely in water.

<h3>What Causes Potassium to Explode in Water?</h3>

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2 years ago
How many moles of silver can be produced from silver nitrate from 1 mole of zinc?
jasenka [17]

Answer:

Answer: 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

Explanation:

To calculate the moles of silver formed and the moles of excess reagent left after the reaction, we need to balance the equation first and need to find the limiting and excess reagent.

The balanced chemical equation is:

Zn + 2AgNO3 ---> Zn (NO3)2 +2Ag

By Stoichiometry:

2 moles of Silver nitrate reacts with 1 mole of Zinc metal

So, 6.5 moles of silver nitrate will react with = 1/2 x 6.5 = 3.25 moles of zinc metal

The required amount of zinc metal is less than the given amount of zinc metal,  hence, it is considered as an excess reagent.

Therefore, silver nitrate is the limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of silver metal

So, 6.5 moles of silver nitrate will produce = 2/2 x 6.5 = 6.5moles of silver metal.

Number of moles of excess reagent left after the completion of reaction = (3.8 - 3.25)moles = 0.55 moles

Hence, 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

4 0
2 years ago
Gaseous methane (CH₄) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O) If 0.3
AveGali [126]

Answer : The percent yield of CO_2 is, 68.4 %

Solution : Given,

Mass of CH_4 = 0.16 g

Mass of O_2 = 0.84 g

Molar mass of CH_4 = 16 g/mole

Molar mass of O_2 = 32 g/mole

Molar mass of CO_2 = 44 g/mole

First we have to calculate the moles of CH_4 and O_2.

\text{ Moles of }CH_4=\frac{\text{ Mass of }CH_4}{\text{ Molar mass of }CH_4}=\frac{0.16g}{16g/mole}=0.01moles

\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{0.84g}{32g/mole}=0.026moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CH_4+2O_2\rightarrow CO_2+2H_2O

From the balanced reaction we conclude that

As, 2 mole of O_2 react with 1 mole of CH_4

So, 0.026 moles of O_2 react with \frac{0.026}{2}=0.013 moles of CH_4

From this we conclude that, CH_4 is an excess reagent because the given moles are greater than the required moles and O_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of CO_2

From the reaction, we conclude that

As, 2 mole of O_2 react to give 1 mole of CO_2

So, 0.026 moles of O_2 react to give \frac{0.026}{2}=0.013 moles of CO_2

Now we have to calculate the mass of CO_2

\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2

\text{ Mass of }CO_2=(0.013moles)\times (44g/mole)=0.572g

Theoretical yield of CO_2 = 0.572 g

Experimental yield of CO_2 = 0.391 g

Now we have to calculate the percent yield of CO_2

\% \text{ yield of }CO_2=\frac{\text{ Experimental yield of }CO_2}{\text{ Theretical yield of }CO_2}\times 100

\% \text{ yield of }CO_2=\frac{0.391g}{0.572g}\times 100=68.4\%

Therefore, the percent yield of CO_2 is, 68.4 %

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