The elements that lack the most properties of metals are called

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$

From the reaction, we conclude that

As, 2 mole of $Na$ react to give 2 mole of $NaCl$

So, 0.217 mole of $HCl$ react to give 0.217 mole of $NaCl$

Now we have to calculate the mass of $NaCl$

$\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl$

Molar mass of $NaCl$ = 58.5 g/mole

$\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g$

Now we have to calculate the percent yield of this reaction.

Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = $\frac{5.24g}{12.7g}\times 100$

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

4 0

3 years ago

What mass of propane could burn in 48.0 g of oxygen? C3H8 + 5O2 → 3CO2 + 4H2O

Ipatiy [6.2K]

Answer:

Mass = 13.23 g

Explanation:

Given data:

Mass of oxygen = 48.0 g

Mass of propane burn = ?

Solution:

Chemical equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 48.0 g/ 32 g/mol

Number of moles = 1.5 mol

now we will compare the moles of propane and oxygen.

O₂ : C₃H₈

5 : 1

1.5 : 1/5×1.5 = 0.3 mol

Mass of propane burn:

Mass = number of moles × molar mass

Mass = 0.3 mol × 44.1 g/mol

Mass = 13.23 g

6 0

2 years ago

Scientists group similar organisms together through a process called …..

Bond [772]

Answer:

plz i need to ask another question i hope u understand

Explanation:

4 0

3 years ago

Read 2 more answers

(C6H6) can be biodegraded by microorganism. if 30 mg of benzene is present, what amount of oxygen required for biodegradation, n

quester [9]

Answer:

36.92 mg of oxygen required for bio-degradation.

Explanation:

$5C_6H_6+15O_2\rightarrow 12CO_2+15H_2O$

Mass of benzene = 30 mg = 0.03 g (1000 mg = 1 g )

Moles benzene = $\frac{0.03 g}{78 g/mol}=0.0003846 mol$

According to reaction 5 moles of benzene reacts with 15 moles of oxygen gas.

Then 0.0003846 mol of benzene will react with:

$\frac{15}{5}\times 0.0003846 mol=0.0011538 mol$ of oxygen gas

Mass of 0.0011538 moles of oxygen gas:

0.0011538 mol × 32 g/mol = 0.03692 g = 36.92 mg

36.92 mg of oxygen required for bio-degradation.

4 0

2 years ago

How many grams of drug powder would you need to weigh to prepare 32ml of a suspension that should have a strength of 20mg/ml?

butalik [34]

15 grams for the drug powder

8 0

2 years ago

Read 2 more answers