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3 years ago

Examine the chemical reaction and lab scenario.

Chemistry

1 answer:

Ierofanga [76]3 years ago

3 0

Explanation:

Percentage Yield

= (3.37g/3.81g) * 100% = 88.45%.

Therefore 88.45% SO2 is the percentage yield.

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How much mass do 3.4 moles of silver contain?

tensa zangetsu [6.8K]

367.2 g of silver

Explanation:

To find the mass of a substance knowing the number of moles we use the following formula:

number of mole = mass / molecular weight

In the case of silver we use the atomic weight of 108 g/mole.

mass = number of moles × molecular weight

mass of silver = 3.4 moles × 108 g/mole

mass of silver = 367.2 g

Learn more about:

moles

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5 0

2 years ago

Different measurements are expressed in different units. Choose the correct Sl units for the lo

swat32

Answer:

The SI units for measuring the velocity of the car: meters per second (m/s)

The Sl units for measuring the acceleration of the car: meters per second squared (m/s^2)

The SI units for measuring force: Newton (N)

The SI units for measuring mass: kilograms (kg)

Explanation:

Système international (SI) or International System of Units consist of a list of unit measurement that mostly used by scientist. The scientist from a different country might use different unit that makes them have to convert the result of each other. The usage of the same unit measurement will help the scientist to read that publication easier, make it easier to share and discuss any topic. The unit used is metric since the decimal system also makes the conversion of a unit easier.

8 0

2 years ago

How many grams of fluorine are needed to generate 7.65 moles Carbon Tetrafluoride

Delicious77 [7]

Answer:

36.3375 g

Explanation:

7 0

2 years ago

Len [333]

A.) AlO is the correct formula

8 0

3 years ago

Find percent yield:

saveliy_v [14]

Answer: The percent yield of the reaction is 91.8 %

Explanation:

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

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

For $B_5H_9$ :

Given mass of $B_5H_9$ = 4.0 g

Molar mass of $B_5H_9$ = 63.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol$

For oxygen gas:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol$

The chemical equation for the reaction of $B_5H_9$ and oxygen gas follows:

$2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O$

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of $B_2H_5$

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{12}\times 0.3125=0.052mol$ of $B_2H_5$

As, given amount of $B_2H_5$ is more than the required amount. So, it is considered as an excess reagent.

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

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of $B_2O_3$

So, 0.3125 moles of oxygen gas will produce = $\frac{5}{12}\times 0.3125=0.130moles$ of water

Now, calculating the mass of $B_2O_3$ from equation 1, we get:

Molar mass of $B_2O_3$ = 69.93 g/mol

Moles of $B_2O_3$ = 0.130 moles

Putting values in equation 1, we get:

$0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g$

To calculate the percentage yield of $B_2O_3$ , we use the equation:

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

Experimental yield of $B_2O_3$ = 8.32 g

Theoretical yield of $B_2O_3$ = 9.052 g

Putting values in above equation, we get:

$\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%$

Hence, the percent yield of the reaction is 91.8 %

6 0

3 years ago