1.) 1.35 g H2 is allowed to react with 10.2 g N2, producing 2.95 g NH3. Part A W

ID: 485095 • Letter: 1

Question

1.) 1.35 g H2 is allowed to react with 10.2 g N2, producing 2.95 g NH3.

Part A

What is the theoretical yield in grams for this reaction under the given conditions?

My answer was 7.61g.

Now, My homework is asking.... What is the percent yield for this reaction under the given conditions?

Express your answer to three significant figures and include the appropriate units.

I need to know what is the appropriate units as well.

2.) If 4.0 L of a 5.2 M solution of SrCl2 is diluted to 50 L , what is the molarity of the diluted solution?

3.) A chemist wants to make 5.0

L of a 0.450 M CaCl2solution.

What mass of CaCl2 (in g) should the chemist use?

4.)

Part A

If 4.0 L of a 5.2 M solution of SrCl2 is diluted to 50 L , what is the molarity of the diluted solution?

Express the molarity to two significant figures.

Li2S(aq)+Co(NO3)2(aq)2LiNO3(aq)+CoS(s)

1.)

3 H2 + N2 ------------> 2 NH3

6.0 g 28 g 34 g

1.35 g 10.2 g ??

here limiting reagent is H2. so product formed based on that.

6.0 g H2 -------------> 34 g NH3

1.35 g H2 -------------> ?? NH3

theoretical yield of NH3 = 1.35 x 34 / 6 = 7.65 g

actual yield = 2.95 g

percent yield = (actual / theoretical ) x 100

= (2.95 / 7.65 ) x 100

percent yield = 38.6 %

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