For the following reaction, 14.7 grams of phosphorus (P_4) are allowed to react

Question

For the following reaction, 14.7 grams of phosphorus (P_4) are allowed to react with 39.0 grams ofchlorine gas. phosphorus (P_4)(s) + chlorine(g) phosphorus trichloride(I) What is the maximum mass of phosphorus trichloride that can be formed? (ANSWER HERE) grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? (ANSWER HERE) grams For the following reaction, 6.74 grams of oxygen gas are mixed with excess sulfur dioxide. The reaction yields 31.0 grams of sulfur trioxide. sulfur dioxide (g) + oxygen (g) sulfur trioxide (g) What is the theoretical yield of sulfur trioxide ? (ANSWER HERE) grams What is the percent yield for this reaction? (ANSWER HERE) % For the following reaction, 0.299 moles of iron are mixed with 0.284 moles of oxygen gas. iron (s) + oxygen (g) iron(II) oxide (s) What is the FORMULA for the limiting reagent? What is the maximum amount of iron(II) oxide that can be formed? ANSWER IN moles

Explanation / Answer

4)

we know that

moles = mass / molar mass

moles of P4 = 14.7 / 123.9 = 0.118644

moles of Cl2 = 39 / 70.91 = 0.55

now

the balanced reaction is

P4 + 6 Cl2 --> 4 PCl3

we can see that

moles of Cl2 required = 6 x moles of P4 = 6 x 0.118644 = 0.712

but only 0.55 moles of Cl2 are present

so

Cl2 is the limiting reagent

now

theoretical moles of PCl3 produced = (4/6) x moles of 02 = 4 x 0.55 / 6 = 0.3666

now

mass = moles x molar mass

so

theoretical mass of PCl3 produced = 0.3666 x 137.33 = 50.35

so

50.35 grams of PCl3 can be produced

5)

we know that

moles = masss / molar mass

so

moles of 02 taken = 6.74 / 32 = 0.21

now

the balanced reaction is

2 S02 + 02 --> 2 S03

we can see that

theoretical moles of S03 = 2 x moles of 02 taken = 2 x 0.21 = 0.42

now

mass = moles x molar mass

theoretical mass of S03 produced = 0.42 x 80 = 33.7

so

the theoretical yield of sulfur trioxide is 33.7 g

b) % yield = actual x 100 / theoretical

% yield = 31 x 100 /33.7

% yield = 91.99

so

the percent yield for this reaction is 91.99 %

6)

the balanced reaction is

2 Fe + 02 --> 2 FeO

we can see that

moles of Fe required = 2 x moles of 02

moles of Fe required = 2 x 0.284

moles of Fe required = 0.568

so

0.568 moles of Fe is needed

but only 0.299 moles of Fe are present

so

Fe is the limiting reagent

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