A group of tissues that work together to perform a specific function is called

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

Thea is doing a chemistry experiment. The instructions say she needs to use pure water. Hamza offer her a bottle labelled “100%

Diano4ka-milaya [45]

How is this going to help us in life wat the fawk

5 0

3 years ago

HELP. I DONT KNOW WHAT TO DO. Barium sulfate, BaSO4, is made by the following reaction.

Ghella [55]

Answer:

$\boxed{\text{66.95 g BaSO$_{4}$}}$

Explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.

M_r: 261.34 233.39

Ba(NO₃)₂ + Na₂SO₄ ⟶ BaSO₄ + 2NaNO₃

m/g: 75.00

1. Moles of Ba(NO₃)₂

$\text{Moles of Ba(NO$_{3})_{2}$} = \text{75.00 g} \times \dfrac{\text{1 mol}}{\text{261.34 g}} = \text{0.286 98 mol}$

2. Moles of BaSO₄

The molar ratio is (1 mol BaSO₄/1 mol Ba(NO₃)₂

$\text{Moles of BaSO$_{4}$}= \text{0.286 98 mol Ba(NO$_{3})_{2}$ } \times \dfrac{\text{1 mol BaSO$_{4}$}}{\text{1 mol Ba(NO$_{3})_{2}$}} = \text{0.286 98 mol BaSO$_{4}$}$

3. Mass of BaSO₄

$\text{Mass of BaSO$_{4}$} = \text{0.286 98 mol BaSO$_{4}$} \times \dfrac{\text{233.39 g BaSO$_{4}$}}{\text{1 mol BaSO$_{4}$}} = \textbf{66.98 g BaSO$_{4}$}\\\\\text{The theoretical yield of barium sulfate is } \boxed{\textbf{66.98 g BaSO$_{4}$}}$