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3 years ago

2. How many moles of HI would be necessary to react with 0.5 moles of Mn? 2Mn + 6HL - 3H, + 2Mnl,

Chemistry

1 answer:

dimulka [17.4K]3 years ago

3 0

Answer:

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Explanation:

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3 years ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of water is pro

MAVERICK [17]

Answer: The percent yield of water is 46.9 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For sulfuric acid:

Given mass of sulfuric acid = 72.6 g (Assuming)

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 1, we get:

$\text{Moles of sulfuric acid}=\frac{72.6g}{98g/mol}=0.741mol$

For NaOH:

Given mass of NaOH = 77 g (Assuming)

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaOH}=\frac{77g}{40g/mol}=1.925mol$

The chemical equation for the reaction of sulfuric acid and sodium hydroxide follows:

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of NaOH

0.741 moles of sulfuric acid will react with = $\frac{2}{1}\times 0.741=1.482mol$ of NaOH

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

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

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of water

0.741 moles of sulfuric acid will react with = $\frac{2}{1}\times 0.741=1.482mol$ of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 1.482 moles

Putting values in equation 1, we get:

$1.482mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.482mol\times 18g/mol)=26.67g$

To calculate the percentage yield of water, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of water = 12.5 g (Assuming)

Theoretical yield of water = 26.67 g

Putting values in above equation, we get:

$\%\text{ yield of water}=\frac{12.5g}{26.67g}\times 100\\\\\% \text{yield of water}=46.9\%$

Hence, the percent yield of water is 46.9 %

3 0

3 years ago

2. How many moles of HI would be necessary to react with 0.5 moles of Mn? 2Mn + 6HL - 3H, + 2Mnl,​

2. How many moles of HI would be necessary to react with 0.5 moles of Mn? 2Mn + 6HL - 3H, + 2Mnl,