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Chapter 4: 4.9 CYL (page 195)

How many NH3 molecules are produced by the reaction of 4 mol of Ca(OH)2 according to the following reaction

\({\left( {N{H_4}} \right)_2}S{O_4} + Ca{\left( {OH} \right)_2} \to 2N{H_3} + CaS{O_4} + 2{H_2}O\)

Short Answer

Expert verified

4.8 x 1024 molecules of NH3 are produced.

Step by step solution

01

Determining the number of moles of NH3

\({\left( {N{H_4}} \right)_2}S{O_4} + Ca{\left( {OH} \right)_2} \to 2N{H_3} + CaS{O_4} + 2{H_2}O\)

From the given balanced reaction, we can see that

1 mol of Ca (OH)2 produces 2 mol of NH3

So, 4 mol of Ca (OH)2 will produce 8 mol of NH3

02

Determining the number of molecules of NH3

1 mol of NH3 contains 6.022 x 1023 molecules of NH3

So, 8 mol of NH3 will contain

8 x 6.022 x 1023= 4.8 x 1024 molecules of NH3

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Most popular questions from this chapter

A 0.00215g sample of polystyrene, a polymer composed of carbon and hydrogen, produced 0.00726g of CO2 0.00148g of H2O in combustion analysis. What is the empirical formula for polystyrene?

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What volume of 0.0105-M HBr solution is required to titrate 125 mL of a 0.0100-MCa(OH)2 solution?

Ca(OH)2(aq) + 2HBr(aq)鉄禖补叠谤2(aq) + 2H2 O(l)

Which is the limiting reactant when 5.00 g of H2 and 10 g of O2 react to form water?

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