The diagram shows a ball that is dropped and allowed to bounce.

when 42.66 grams of PCl5 react with excess P4O10 the amount of product formed is 47.22 grams of POCl3. What is the percent yield

icang [17]

Percent yield or yield is mathematically defined as:

Yield = Actual amount / Theoretical amount

So to solve the yield, let us first calculate the theoretical amount of POCl3 produced. The balanced chemical reaction for this is:

6 PCl5 + P4O10 ---> 10 POCl3

Since P4O10 is stated to be supplied in large amount, then PCl5 becomes the limiting reactant.

So we calculate for POCl3 based on PCl5. To do this let us convert the amount into moles: (molar mass PCl5 = 208.24 g/mol)

n PCl5 = 42.66 grams / (208.24 g/mol)

n PCl5 = 0.205 mol

Now based on the stoichiometric ratio of the reaction:

n POCl3 = 0.205 mol (10 POCl3 / 6 PCl5)
n POCl3 = 0.3414 mol POCl3

Converting to mass (molar mass POCl3 = 153.33 g/mol)

m POCl3 = 0.3414 mol (153.33 g/mol)

m POCl3 = 52.35 g

Calculating for yield:

Yield = 47.22 g/ 52.35 g

Yield = 0.902

%Yield = 90.2 % (ANSWER)

6 0

3 years ago

Zinc Mass If a 1.85 g mass of zinc produces 475 mL of gas and your balloon weighs 0.580 g and the room temperature is 21.5°C. Ca

Alexus [3.1K]

The weight of the balloon is irrelevant because it is the gas that lifts it in the air. We are already given with the required volume, so we use this instead. The atomic weight of zinc is 65.38 g/mol. Assuming ideal gas behavior,

PV=nRT
P(475 mL)(1 L/1000 mL) = (1.85/65.38)(0.0821 L·atm/mol·K)(21.5 + 273)
P = 1.44 atm

Then, we use this pressure and the volume to find the moles of zinc.

(1.44 atm)(475 mL+1 mL)(1 L/1000 mL) = n(0.0821 L·atm/mol·K)(21.5 + 273)
Solving for n,
n = 0.02836 moles of zinc

3 0

4 years ago

How many milliliters of a 0.40%(w/v) solution of nalorphine must be injected to obtain a dose of 1.5 mg?

ozzi

The number of Ml of a 0.40 %w/v solution of ,nalorphine that must be injected to obtain a dose of 1.5 mg is calculated as below

since M/v% is mass of solute in grams per 100 ml

convert Mg to g
1 g = 1000 mg what about 1.5 mg =? grams
= 1.5 /1000 = 0.0015 grams

volume is therefore = 100 ( mass/ M/v%)

= 100 x( 0.0015/ 0.4) = 0.375 ML

6 0

4 years ago

A 34.0 gram sample of an un-known hydrocarbon is burned in excess oxygen to form 93.5 grams of carbon dioxide and 76.5 grams of

natka813 [3]

Answer:

The answer to your question is CH₄

Explanation:

Process

1.- Write the possible chemical reaction

CxHy + O₂ ⇒ CO₂ + H₂O

2.- Calculate the amount of carbon in the reaction

Molecular mass of CO₂ = 12 + 32 = 44 g

44 g of CO₂ ------------------- 12 g of carbon

93.5 g of CO₂ ---------------- x

x = (93.5 x 12) / 44

x = 25.5 g of carbon

3.- Calculate the mass of hydrogen in the sample

Molecular mass of water = 2 + 16 = 18 g

18 g of water ------------- 2 g of hydrogen

76.5 g of water -------- x g of hydrogen

x = (76.5 x 2) / 18

x = 8.5 g of hydrogen

4.- Calculate the moles of carbon and hydrogen in the sample

12g of carbon ------------- 1 mol

25.5g of carbon ---------- x

x = (25.5 x 1) / 12

x = 2.12 moles of carbon

1 g of hydrogen ----------- 1 mol

8.5 g of hydrogen -------- x

x = (8.5 x 1) / 1

x = 8.5 moles of hydrogen

5.- Divide both number of moles by the lowest number

carbon = 2.12 / 2.12 = 1

hydrogen = 2.12 = 8,5 / 2.12 = 4

6.- Write the molecular formula

CH₄

7 0

3 years ago

Read 2 more answers

The ksp of calcium carbonate, caco3, is 3.36 × 10-9 m2. calculate the solubility of this compound in g/l.

maw [93]

CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X

Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²

Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M

Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L

Molar mass of CaCO₃ = 100 g mol⁻¹

Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L

7 0

3 years ago