Determine the volume (L) of nitrogen monoxide gas that is created at STP when 32.2 g

of solid copper reacts with excess nitric acid.

3Cu(s) + 8HNO3(aq) — 3Cu(NO3)2 (aq) + 4H2O(1) + 2NO(g)

Answers

Answer 1

Taking into account the reaction stoichiometry and STP conditions, the volume of nitrogen monoxide gas that is created at STP when 32.2 g of solid copper reacts with excess nitric acid is 7.5677 L.

Reaction stoichiometry

The balanced reaction is:

3 Cu(s) + 8 HNO₃(aq) → 3 Cu(NO₃)₂ (aq) + 4 H₂O(l) + 2 NO(g)

By reaction stoichiometry, the following amounts of moles of each compound participate in the reaction:

Cu: 3 molesHNO₃: 8 molesCu(NO₃)₂: 3 molesH₂O: 4 moles NO: 2 moles

The molar mass of the compounds is:

Cu: 63.54 g/moleHNO₃: 63 g/moleCu(NO₃)₂: 187.54 g/moleH₂O: 18 g/moleNO: 30 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Cu: 3 moles× 63.54 g/mole= 190.62 gramsHNO₃: 8 moles× 63 g/mole= 504 gramsCu(NO₃)₂: 3 moles ×187.54 g/mole= 562.62 gramsH₂O: 4 moles ×18 g/mole= 72 gramsNO: 2 moles ×30 g/mole= 60 grams

STP conditions

The STP conditions refer to the standard temperature and pressure, which values are 0 °C and 1 atmosphere and are reference values for gases. In these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

Moles of NO formed

The following rule of three can be applied: if by reaction stoichiometry 190.62 grams of Cu form 2 moles of NO, 32.2 grams of Cu form how many moles of NO?

moles of NO= (32.2 grams of Cu× 2 moles of NO)÷ 190.62 grams of Cu

moles of NO= 0.3378 moles

Then, 0.3378 moles of NO are formed.

Volume of NO created

Now, you can apply the following rule of three: if by definition of STP conditions 1 mole of NO occupies a volume of 22.4 liters, 0.3378 moles occupies how much volume?

volume= (0.3378 moles× 22.4 L)÷ 1 mole

volume= 7.5677 L

Finally, the volume of NO created is 7.5677 L.

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Answer:

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Answer:

Option (3) 11

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Answer:

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Answers

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Explanation:

A 0.500-g sample containing Na2CO3 plus inert matter is analyzed by adding 50.0 mL of 0.100 M HCl, a slight excess, boiling to remove CO2, and then back-titrating the excess acid with 0.100 M NaOH. If 5.6 mL NaOH is required for the back-titration, what is the percent Na2CO3 in the sample?

Answers

The percentage by mass of Na2CO3 in the sample is 48%.

The equation of the reaction of Na2CO3 with HCl;

Na2CO3(aq) + 2HCl(aq) ------> 2NaCl(aq) + H2O(l) + CO2(g)

Since the HCl is in excess, the excess is back titrated with NaOH as follows;

NaOH (aq) + HCl(aq) ---->NaCl(aq) + H2O(l)

Number of moles of HCl added =  0.100 M × 50/1000 L = 0.005 moles

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Lithium has an atomic mass of 6.941 amu. Lithium has two isotopes. One isotopes has a mass of 6.015 amu with a relative abundance of 7.49%. What is the mass of the other isotopes?

Answers

Explanation:

List the known and unknown quantities and plan the problem. Change each percent abundance into decimal form by dividing by 100. Multiply this value by the atomic mass of that isotope. Add together for each isotope to get the average atomic mass.

Answer:

The atomic mass of second isotope is 7.016

Explanation:

Given data:

Average Atomic mass of lithium = 6.941 amu

Atomic mass of first isotope = 6.015 amu

Relative abundance of first isotope = 7.49%

Abundance of second isotope = ?

Atomic mass of other isotope = ?

Solution:

Total abundance = 100%

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Now we will calculate the mass if second isotope.

Average atomic mass of lithium = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

6.941 = (6.015×7.49)+(x×92.51) /100

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6.941×100 = 45.05235 + (x92.51)

694.1 - 45.05235   = (x92.51)

649.04765 = x 92.51

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