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Chapter 3: Problem 134

Consider the following unbalanced reaction: $$ \mathrm{P}_{4}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{PF}_{3}(g) $$ What mass of \(\mathrm{F}_{2}\) is needed to produce \(120 . \mathrm{g}\) of \(\mathrm{PF}_{3}\) if the reaction has a 78.1\(\%\) yield?

Short Answer

Expert verified
99.25 g of F2 is needed to produce 120 g of PF3 with a 78.1% reaction yield.

Step by step solution

01

Balance the chemical equation

First, let's balance the given chemical equation: \(P_4(s) + 6F_2(g) \longrightarrow 4PF_3(g)\)
02

Calculate the amount of PF3 in moles

To calculate the amount of PF3 produced, we will convert the given mass (120 g) to moles. The molar mass of PF3 can be found using the periodic table: \(1P: 30.97g/mol\) \(3F: 19.00g/mol * 3 = 57.00g/mol\) Adding these together, the molar mass of PF3 is 87.97 g/mol. Now, let's convert the given mass of PF3 to moles: \(\frac{120\,\mathrm{g}}{1} * \frac{1\,\mathrm{mol\,PF_3}}{87.97\,\mathrm{g}} = 1.36\,\mathrm{mol\,PF_3}\)
03

Determine the amount of F2 required

Utilizing the stoichiometric coefficients, we can determine how many moles of F2 are required to produce the calculated moles of PF3. The mole ratio of F2 to PF3 is 6:4, which simplifies to 3:2 \(\frac{1.36\,\mathrm{mol\,PF_3}}{1} * \frac{3\,\mathrm{mol\,F_2}}{2\,\mathrm{mol\,PF_3}} = 2.04\,\mathrm{mol\,F_2}\)
04

Convert moles of F2 to mass

Now we need to convert the moles of F2 required into mass. The molar mass of F2 is: \(2F: 19.00g/mol * 2 = 38.00g/mol\) Converting moles of F2 to mass: \(2.04\,\mathrm{mol\,F_2} * \frac{38.00\,\mathrm{g}}{1\,\mathrm{mol\,F_2}} = 77.52\,\mathrm{g\,F_2}\)
05

Adjust the mass of F2 needed based on the reaction yield

Now we need to take into account the reaction yield which is 78.1%. This means that the actual mass of F2 consumed is greater than what we calculated in step 4. To find the mass of F2 required, we will divide the mass calculated in step 4 by the reaction yield: \(\frac{77.52\,\mathrm{g\,F_2}}{0.781} = 99.25\,\mathrm{g\,F_2}\) Therefore, 99.25 g of F2 is needed to produce 120 g of PF3 with a 78.1% reaction yield.

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Most popular questions from this chapter

Consider the following unbalanced equation: $$ \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{CaSO}_{4}(s)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) $$ What masses of calcium sulfate and phosphoric acid can be produced from the reaction of 1.0 \(\mathrm{kg}\) calcium phosphate with 1.0 \(\mathrm{kg}\) concentrated sulfuric acid $\left(98 \% \mathrm{H}_{2} \mathrm{SO}_{4} \text { by mass)? }\right.$

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In the production of printed circuit boards for the electronics industry, a 0.60 -mm layer of copper is laminated onto an insulating plastic board. Next, a circuit pattern made of a chemically resistant polymer is printed on the board. The unwanted copper is removed by chemical etching, and the protective polymer is finally removed by solvents. One etching reaction is $$ \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q)+\mathrm{Cu}(s) \longrightarrow $$ $$ 2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q) $$ A plant needs to manufacture \(10,000\) printed circuit boards, each $8.0 \times 16.0 \mathrm{cm}\( in area. An average of \)80 . \%$ of the copper is removed from each board (density of copper \(=8.96 \mathrm{g} / \mathrm{cm}^{3}\) . What masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) are needed to do this? Assume 100\(\%\) yield.

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