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Anna71 [15]
3 years ago
14

Ammonium phosphate is an important ingredient in many solid fertilizers. it can be made by reacting aqueous phosphoric acid with

liquid ammonia. calculate the moles of ammonium phosphate produced by the reaction of 0.085 mol of ammonia. be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
Chemistry
1 answer:
Debora [2.8K]3 years ago
3 0

Answer:

0.028 mole of ammonium phosphate, (NH₄)₃PO₄.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

H₃PO₄ + 3NH₃ —> (NH₄)₃PO₄

From the balanced equation above,

3 moles of NH₃ reacted to produce 1 mole of (NH₄)₃PO₄.

Finally, we shall determine the number of mole of (NH₄)₃PO₄ produced by the reaction of 0.085 mole of ammonia, NH₃. This can be obtained as follow:

From the balanced equation above,

3 moles of NH₃ reacted to produce 1 mole of (NH₄)₃PO₄.

Therefore, 0.085 mole of NH₃ will react to produce = (0.085 × 1)/3 = 0.028 mole of (NH₄)₃PO₄.

Thus, 0.028 mole of ammonium phosphate, (NH₄)₃PO₄ were obtained from the reaction.

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A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor
AlexFokin [52]

Answer : The partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of C_2HBrClF_3 = 15.0 g

Mass of O_2 = 22.6 g

Molar mass of C_2HBrClF_3 = 197.4 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_2HBrClF_3 and O_2.

\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole

and,

\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole

Now we have to calculate the mole fraction of C_2HBrClF_3 and O_2.

\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971

and,

\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903

Now we have to partial pressure of C_2HBrClF_3 and O_2.

According to the Raoult's law,

p^o=X\times p_T

where,

p^o = partial pressure of gas

p_T = total pressure of gas

X = mole fraction of gas

p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T

p_{C_2HBrClF_3}=0.0971\times 862torr=84torr

and,

p_{O_2}=X_{O_2}\times p_T

p_{O_2}=0.903\times 862torr=778torr

Therefore, the partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

6 0
3 years ago
William measures a test tube and finds that the mass of the test tube is 5g. He places the lone reactant in the test tube and fi
iogann1982 [59]

86 percent is the percent yield for this experiment if he expected to produce 5g of product.

Explanation:

Given that:

mass of test tube = 5 grams

mass of test tube + reactant is 12.5 grams

mass of reactant = ( mass of test tube + reactant ) - (mass of test tube)

mass of reactant = 12.5 -5

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when 7.5 grams of reactant is heated mass of test tube was found to be 9.3 grams.

so mass of product formed = 9.3 - 5

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However, he expected the product to be 5 grams (theoretical yield)

Percent yield = \frac{actual yield}{theoretical yield} x 100

          putting the values in the formula:

percent yield = \frac{4.3}{5} x 100

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Answer: Option (B) is the correct answer.

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This continuous arrangement of atoms are like a lattice.

For example, diamond is a covalent network solid in which carbon atoms are arranged in a continuous lattice like structure.

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