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3 years ago

Please help! Thanks

Chemistry

1 answer:

trapecia [35]3 years ago

7 0

Th actual yield of the reaction is 24.86 g

We'll begin by calculating the theoretical yield of the reaction.

2Na + Cl₂ → 2NaCl

Molar mass of Na = 23 g/mol

Mass of Na from the balanced equation = 2 × 23 = 46 g

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl from the balanced = 2 × 58.5 = 117 g

From the balanced equation above,

46 g of Na reacted to produce 117 g of NaCl.

Therefore,

11.5 g of Na will react to produce = (11.5 × 117) / 46 = 29.25 g of NaCl.

Thus, the theoretical yield of NaCl is 29.25 g.

Finally, we shall determine the actual yield of NaCl.

Percentage yield = 85%

Theoretical yield = 29.25 g

Actual yield =?

Actual yield = Percent yield × Theoretical yield

Actual yield = 85% × 29.25

Actual yield = 0.85 × 29.25 g

Actual yield = 24.86 g

Learn more about stoichiometry: brainly.com/question/25899385

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A gas at 928 kpa, 129 C occupies a volume of 569 L. Calculate the volume at 319 kpa and 32 C.

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Answer:

1255.4L

Explanation:

Given parameters:

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Unknown:

V₂ = ?

Solution:

The combined gas law application to this problem can help us solve it. It is mathematically expressed as;

$\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }$

P, V and T are pressure, volume and temperature

where 1 and 2 are initial and final states.

Now,

take the units to the appropriate ones;

kpa to atm, °C to K

P₂ = 319kpa in atm gives 3.15atm

P₁ = 928kpa gives 9.16atm

T₂ = 32°C gives 273 + 32 = 305K

T₁ = 129°C gives 129 + 273 = 402K

Input the values in the equation and solve for V₂;

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Read 2 more answers

What mass of nitrogen is needed to fill an 855 L tank at STP?

scoundrel [369]

Answer:

It takes 1,068.76 grams of nitrogen to fill an 855 L tank at STP.

Explanation:

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C or 273.15 °K are used and are reference values for gases.

On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

So, in this case:

P= 1 atm
V= 855 L
n= ?
R= 0.082 $\frac{atm*L}{mol*K}$
T= 273.15 K

Replacing:

1 atm* 855 L= n* 0.082 $\frac{atm*L}{mol*K}$ * 273.15 K

Solving:

$n=\frac{1 atm* 855 L}{0.082\frac{atm*L}{mol*K} *273.15 K }$

n= 38.17 moles

Being the molar mass of nitrogen N2 equal to 28 g / mol, you can apply the following rule of three: if there are 28 grams in 1 mole, how much mass is there in 38.17 moles?

$mass=\frac{38.17 moles*28 grams}{1 mole}$

mass= 1,068.76 grams

It takes 1,068.76 grams of nitrogen to fill an 855 L tank at STP.

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