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2 years ago

Calculate the theoretical yield if 100.0 g p4o10 react with 200.0 g h2o

1 answer:

3 0

The reaction of Phosphorous Pentaoxide with water yield Phosphoric Acid as shown below,

P₄O₁₀ + 6 H₂O → 4 H₃PO₄

According to balance equation,

283.88 g (1 mole) P₄O₁₀ requires = 108 g (6 mole) of H₂O
So,
100 g P₄O₁₀ will require = X g of H₂O

Solving for X,

X = (100 g × 108 g) ÷ 283.88 g

X = 38.04 g of H₂O

So, 100 g P₄O₁₀ requires 38.04 g of H₂O, while we are provided with 200 g of H₂O which means that water is in excess and P₄O₁₀ is limiting reagent. Therefore, P₄O₁₀ will control the yield of H₃PO₄. So,
As,
283.88 g (1 mole) P₄O₁₀ produced = 391.96 g (4 mole) of H₃PO₄
So,
100 g P₄O₁₀ will produce = X g of H₃PO₄

Solving for X,
X = (100 g × 391.96 g) ÷ 283.88 g

X = 138.07 g of H₃PO₄

Result:
Theoretical Yield of this reaction is 138.07 g.

You might be interested in

calculate the specific heat capacity for gold n 105 joules are required to heat 30.0 grams of gold from 27.7c to 54.9c

Pepsi [2]

Answer:

The specific heat capacity for gold in 105 joules which are required to heat 30.0 grams of gold is 0.129 J/(g℃)

Explanation:

We make use of the formula

$Q=m \times c \times \Delta T$

where

∆T = final T - initial T

= 54.9℃ - 27.7℃ = 27.2℃

Q is the heat energy in Joules = 105J

c is the specific heat capacity = ?

m is the mass of Gold = 30.0g

$Q=m \times c \times \Delta T$

Rearranging the formula

$c= \frac {Q}{(m\times \Delta T)}$

$= \frac {105J}{(30.0g \times 27.2 ^\circ{C})}\\\\= \frac {105J}{(816g^\circ{C})}$

So,

c = 0.129 J/(g℃)

(Answer)

7 0

2 years ago

Calculate the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 3

ZanzabumX [31]

Answer:

19,26 kJ

Explanation:

The work done when a gas expand with a constant atmospheric pressure is:

W = PΔV

Where P is pressure and ΔV is the change in volume of gas.

Assuming the initial volume is 0, the reaction of 500g of Zn with H⁺ (Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)) produce:

500,0g Zn(s)× $\frac{1molZn}{65,38g}$ × $\frac{1molH_{2}(g)}{1molZn}$ = 7,648 moles of H₂

At 1,00atm and 303,15K (30°C), the volume of these moles of gas is:

V = nRT/P

V = 7,648mol×0,082atmL/molK×303,15K / 1,00atm

V = 190,1L

That means that ΔV is:

190,1L - 0L = 190,1L

And the work done is:

W = 1atm×190,1L = 190,1atmL.

In joules:

190,1 atmL× $\frac{101,325}{1atmL}$ = 19,26 kJ

I hope it helps!

5 0

2 years ago

A tanker truck carrying 6.05×103 kg of concentrated sulfuric acid solution tips over and spills its load. The sulfuric acid solu

irinina [24]

Answer:

6,216.684 kilograms of sodium carbonate must be added to neutralize $6.05\times 10^3 kg$ of sulfuric acid solution.

Explanation:

Mass of sulfuric acid solution = $6.05\times 10^3 kg=6.05\times 10^6 g$

$1 kg = 10^3 g$

Percentage mass of sulfuric acid = 95.0%

Mass of sulfuric acid = $\frac{95.0}{100}\times 6.05\times 10^6 g$

$=5,747,500 g$

Moles of sulfuric acid = $\frac{5,747,500 g}{98 g/mol}=58,647.96 mol$

$H_2SO_4+Na_2CO_3\rightarrow Na_2SO_4+CO_2+H_2O$

According to reaction , 1 mole of sulfuric acid is neutralized by 1 mole of sodium carbonate.

Then 58,647.96 moles of sulfuric acisd will be neutralized by :

$\frac{1}{1}\times 58,647.96 mol=58,647.96 mol$ of sodium carbonate

Mass of 58,647.96 moles of sodium carbonate :

$106 g/mol\times 58,647.96 mol=6,216,683.76 g$

6,216,683.76 g = 6,216,683.76 × 0.001 kg = 6,216.684 kg

6,216.684 kilograms of sodium carbonate must be added to neutralize $6.05\times 10^3 kg$ of sulfuric acid solution.

3 0

2 years ago

Which sentence is a scientific statement? A. There is a substance around us that cannot be detected. B. Kids should go to school

stellarik [79]

Answer:

Experimentation is the best way to obtain knowledge.

Explanation:

This is a statement and it is scientific

6 0

2 years ago

Which of the reagents listed below would efficiently accomplish the transformation of 2-methyl-3-cyclopentenol into 2-methyl-3-c

Anni [7]

The answer to this question would be CrO3 or H2SO4. These reagents would efficiently accomplish the transformation of 2-methyl-3-cyclopentenol into 2-methyl-3-cyclopentenone. The synthetic strategy is used for the bond forming and transformation of the molecule. The reagent required depends on the type of molecule.

5 0

2 years ago