1. What volume of 0.240M H2SO4 is needed to react with 42.0mL of 0.120M NaOH? Th

Question

1. What volume of 0.240M H2SO4 is needed to react with 42.0mL of 0.120M NaOH? The equation is

H2SO4(aq)+2NaOH(aq)?Na2SO4(aq)+2H2O(l)

2. What is the molarity of an HNO3 solution if 57.0mL is needed to react with 25.0 mL of 0.150 MKOHsolution? The equation is
HNO3(aq)+KOH(aq)?KNO3(aq)+H2O(l)

3. Assume that you have 1.43mol of H2 and 3.42mol of N2. How many grams of ammonia (NH3)can you make, and how many grams of which reactant will be left over?
3H2+N2?2NH3

Part A: m NH3 =

Part B: m H2 =

Part C: m N2=

I dont understand what how to get these please show step by step!!! thank you

Explanation / Answer

1 ) we know that

N1V1 = N2V2 (milliequivalents of base = milliequivalents of acid)

Normality = Molarity X 2 ( for dibasic acid)

so, 0.250 X 2 X x = 42 X 0.120

x = 10.5 mL

2)

M1V1 = M2V2

M X 57 = 25 X 0.150

Molarity = 0.065 M

3) as form equation

3 moles of hydrogen reacts with one mole of nitrogen to produce 2 moles of NH3

so with, 1.43 moles of hydrogen will react with 1/3 moles of nitrogen ( 1 X 1.43 / 3)

amount of nitrogen reacted with 1.43 moles of hydrogen = 0.477 moles

as given, 1 mole of nitrogen reacts to form 2 moles of ammonia

so, 0.477 moles of nitrogen will react to give 2 X 0.477 moles of ammonia

2 X 0.477 moles = 0.954 moles = 16.218 grams of ammonia

mass of hydrogen left = 0 gm

moles of nitrogen left = 2.943

,mass of nitrogen left = 82.404 g

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