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Tatiana [17]
3 years ago
7

what+is+the+theoretical+yield+of+ammonia(in+grams)+of+19.25+grams+of+nitrogen+gas+and+11.35+grams+of+hydrogen+gas+are+allowed+to

+react
Chemistry
1 answer:
kondor19780726 [428]3 years ago
7 0
We are given the amount of Nitrogen gas and hydrogen gas reacted to form ammonia:

N2 = 19.25 grams
H2 = 11.35 grams

Set-up a balanced chemical equation:

N2 + 3H2 ==> 2NH3

The theoretical amount of ammonia that will be produced from the given amounts is:

First, we need to determine the limiting reactant to serve as our basis for calculation.

number of moles / stoichiometric ratio 

N2 = 19.25  g/ 28 g/mol / 1 = 0.6875
H2 = 11.35 g/ 2 g/mol /3 = 1.89 

The limiting reactant is N2.

0.6875 moles N2 * (2 NH3/ 1 N2) * 17 g/mol NH3

The amount of NH3 produced is 23.375 grams of ammonia. <span />
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Really darn cold lol
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The benzoate ion, c6h5coo− is a weak base with kb=1.6×10−10. how many moles of sodium benzoate are present in 0.50 l of a soluti
lyudmila [28]

NaC6H5COO \rightarrow Na{^{+}} + C6H5COO^{-}

Here the base is a benzoate ion, which is a weak base and reacts with water.

C6H5COO^{-}(aq) + H2O (l)\leftrightarrow C6H5COOH(aq)+ OH^{-}(aq)

The equation indicates that for every mole of OH- that is produced , there is one mole of C6H5COOH produced.

Therefore [OH-] = [C6H5COOH]

In the question value of PH is given and by using pH we can calculate pOH and then using pOH we can calculate [OH-]

pOH = 14 - pH

pH given = 9.04

pOH = 14-9.04 = 4.96

pOH = -log[OH-] or [OH^{-}] = 10^{^{-pOH}}

[OH^{-}] = 10^{^{-4.96}}

[OH^{-}] = 1.1\times 10^{-5}

The base dissociation equation kb = \frac{Product}{Reactant}

kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}

H2O(l) is not included in the 'kb' equation because 'solid' and 'liquid' are taken as unity that is 1.

Value of Kb is given = 1.6\times 10^{-10}

And value of [OH-] we have calculated as 1.1\times 10^{-5} and value of C6H5COOH is equal to OH-

Now putting the values in the 'kb' equation we can find the concentration of C6H5COO-

kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}

1.6\times 10^{-10} = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{[C6H5COO^{-}]}

[C6H5COO^{-}] = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{1.6\times 10^{-10}}

[C6H5COO^{-}] = 0.76 M or 0.76\frac{mol}{L}

So, Concentration of NaC6H5COO would also be 0.76 M and volume is given to us 0.50 L , now moles can we calculated as : Moles = M X L

Moles of NaC6H5COO would be = 0.76(\frac{mol}{L}) \times (0.50L)

Moles of NaC6H5COO (sodium benzoate) = 0.38 mol

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Using Charles law if the volume of a gas is 5.0 L when the temperature is 5.0C what will the new volume be if the temperature is
Vikentia [17]

Answer:

The answer to your question is   V2 = 5.09 l

Explanation:

Data

Volume 1 = V1 = 5.0 L

Temperature 1 = T1 = 5°C

Volume 2 = V2 = ?

Temperature 2 = T2 = 10°C

Formula (Charles law)

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-Solve for V2

                  V2 = V1T2 / T1

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T1 = 5 + 273 = 278°K

T2 = 10 + 273 = 283°K

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                 V2 = (5)(283) / (278)

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