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4 years ago

Consider the balanced equation for the following reaction:

Chemistry

1 answer:

mafiozo [28]4 years ago

3 0

Answer:

There will remain 78.9 grams of NH3.

Explanation:

Step 1: Data given

Mass CuO = 17.3 grams

Molar mass CuO = 79.545 g/mol

Mass NH3 = 81.4 grams

Molar mass NH3 = 17.03 g/mol

Step 2: The balanced equation

3CuO(s) + 2NH3(g) → 3H2O(l) + 3Cu(s) + N2(g)

Step 3: Calculate moles CuO

Moles CuO = mass CuO/ molar mass CuO

Moles CuO = 17.3 grams / 79.545 g/mol

Moles CuO = 0.217 moles

Step 4: Calculate moles NH3

Moles NH3 = 81.4 grams / 17.03 g/mol

Moles NH3 = 4.78 moles

Step 5: Calculate the limiting reactant

CuO is the limiting reactant. It will completely react. NH3 is in excess. There will react 2/3 * 0.217 = 0.145 moles. There will remain 4.78 - 0.145 = 4.635 moles NH3

Step 6: Calculate mass NH3 remaining

Mass NH3 = 4.635 moles * 17.03 g/mol

Mass NH3 = 78.9 grams

There will remain 78.9 grams of NH3.

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3 years ago

An aqueous solution is listed as being 33.8% solute by mass with a density of 1.15 g/mL, the molar mass of the solute is 145.6 g

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Answer:

A) 2.69 M

B) 0.059

Explanation:

A) We have:

33.8% solute by mass= 33.8 g solute/100 g solution

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molarity= $\frac{33.8 g solute}{100 g solution}$ x $\frac{1.15 g solution}{1 ml}$ x $\frac{1 mol solute}{145.6 g solute}$ x $\frac{1000 ml}{1 L}$

molarity= 2.69 mol solute/L solution = 2.69 M

B) We know that there are 33.8 g of solute in 100 g of solution.

As the total solution is compounded by solute+solvent (in this case, solvent is water), the mass of water is the difference between the mass of the total solution and the mass of solute:

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Now, we calculate the number of mol of both solute and water:

mol solute= 33.8 g solute x $\frac{1 mol solute}{145.6 g}$ = 0.232 mol

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Finally, the mol fraction of solute (Xsolute) is calculated as follows:

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3 years ago

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