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Chapter 3: Problem 77

The fertilizer ammonium sulfate \(\left[\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\right]\) is prepared by the reaction between ammonia \(\left(\mathrm{NH}_{3}\right)\) and sulfuric acid: $$ 2 \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(a q) $$ How many kilograms of \(\mathrm{NH}_{3}\) are needed to produce \(1.00 \times 10^{5} \mathrm{~kg}\) of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} ?\)

Short Answer

Expert verified
It requires approximately \(1.28 \times 10^{6} kg\) of Ammonia to produce \(1.00 \times 10^{5} kg\) of Ammonium Sulfate.

Step by step solution

01

Compute the Molar Mass of Ammonium Sulfate and Ammonia

The molar mass of Ammonium Sulfate \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) and Ammonia \(\mathrm{NH}_{3}\) are to be computed. The molar mass of Ammonium Sulfate is \(2(14.01 g/mol + 1.008 g/mol + 15.999 g/mol) + 32.07 g/mol + 4(15.999 g/mol) = 132.14 g/mol\). The molar mass of Ammonia is \(14.01 g/mol + 3(1.008 g/mol) = 17.03 g/mol\).
02

Convert Mass of Ammonium Sulfate to Moles

Using the calculated molar mass of Ammonium Sulfate, convert the given mass of Ammonium Sulfate to moles. This is done using the formula \(moles = \frac{mass}{molar mass}\). Hence, \(\frac{1.00 \times 10^{5} kg}{132.14 g/mol}\) is calculated. Note that \(1 kg = 1000 g\), so the mass of Ammonium Sulfate can be converted to grams.
03

Apply Stoichiometry

From the balanced chemical reaction, it could be observed that 2 moles of Ammonia react to form 1 mole of Ammonium Sulfate. Thus, the amount of moles of Ammonia required would be twice the amount of moles of Ammonium Sulfate calculated. Multiply the moles of Ammonium Sulfate by 2.
04

Convert Moles of Ammonia to Mass

Using the molar mass of Ammonia, convert the moles of Ammonia back to mass using the formula \(mass = moles \times molar mass\). Calculate the mass by multiplying the moles gotten from step 3 by the molar mass of Ammonia. The answer will be in gram, but it can be converted back to kilogram by dividing it by 1000.

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Most popular questions from this chapter

Calculate the number of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) atoms in \(1.50 \mathrm{~g}\) of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\), a sugar.

Nitric oxide (NO) reacts with oxygen gas to form nitrogen dioxide \(\left(\mathrm{NO}_{2}\right),\) a dark-brown gas: $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ In one experiment 0.886 mole of \(\mathrm{NO}\) is mixed with 0.503 mole of \(\mathrm{O}_{2}\). Calculate which of the two reactants is the limiting reagent. Calculate also the number of moles of \(\mathrm{NO}_{2}\) produced.

An impure sample of zinc \((\mathrm{Zn})\) is treated with an excess of sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)\) to form zinc sulfate \(\left(\mathrm{ZnSO}_{4}\right)\) and molecular hydrogen \(\left(\mathrm{H}_{2}\right) .\) (a) Write a balanced equation for the reaction. (b) If \(0.0764 \mathrm{~g}\) of \(\mathrm{H}_{2}\) is obtained from \(3.86 \mathrm{~g}\) of the sample, calculate the percent purity of the sample. (c) What assumptions must you make in (b)?

A certain metal M forms a bromide containing 53.79 percent Br by mass. What is the chemical formula of the compound?

Describe the steps involved in the mole method.
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