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3 years ago

How many moles of h2o are produced when 3.25 moles of o2 react in 2c2h6 + 7o2 ---> 4co2 + 6h2o

Chemistry

1 answer:

Alenkinab [10]3 years ago

5 0

Answer:

2.78 moles of water are produced.

Explanation:

Given data:

Number of moles of H₂O produced = ?

Number of moles of oxygen react = 3.25 mol

Solution:

Chemical equation:

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Now we will compare the moles of water with oxygen.

O₂ : H₂O

7 : 6

3.25 : 6/7×3.25 = 2.78 mol

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What is the density of an object that has a mass of 6.5 g and when placed in water displaces the volume from 4.5mL to 11.8mL? ro

KengaRu [80]

Answer:

$\rho =0.9g/mL$

Explanation:

Hello,

In this case, due to the volume displacement caused the by the object's submersion, it's volume is:

$V=11.8mL-4.5mL=7.3 mL$

In such a way, considering the mathematical definition of density, it turns out:

$\rho =\frac{m}{V}=\frac{6.5g}{7.3mL}\\ \\\rho =0.89g/mL$

Rounding to the nearest tenth we finally obtain:

$\rho =0.9g/mL$

Regards.

3 0

3 years ago

A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?

Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

$\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles$

Now we have to calculate the moles of $Br_2$

${\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles$

${\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles$

The balanced chemical reaction is,

$2Na(s)+Br_2(g)\rightarrow 2NaBr$

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium bromide

Now we have to calculate the percent yield of reaction

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%$

Therefore, the percent yield is, 93.4%

3 0

3 years ago

Can you please help me

rosijanka [135]

Shure what you need help with

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Answer:

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Explanation:

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3 years ago

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3 years ago