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Unit: Stoichiometry
"Multi-Step Problems" â WS #3
Directions: Use dimensional analysis to perform the following calculations. Be sure to show all of your work, include your units, and round your answers to the correct number of significant figures. How many formula units of sodium nitrate will react with 20.0 g of calcium carbonate to product sodium carbonate and calcium nitrate according to the following reaction?
2 Na(NO3) + Ca(CO3) à Na2(CO3) + Ca(NO3)2
What volume of oxygen is produced at STP when 6.58 x 1024 molecules of water is decomposed according to the following reaction? 2 H2O à 2 H2 + O2
How many liters of nitrogen gas are needed to react with 90.0 g of potassium at STP in order to produce potassium nitride according to the following reaction? 6 K + N2 à 2 K3N

Respuesta :

mickymike92

Answer:

1. 2.41 × 1023 formula units

2. 122 L

3. 7.81 L

Explanation:

1. Equation of the reaction: 2 Na(NO3) + Ca(CO3) ---> Na2(CO3) + Ca(NO3)2

Mole ratio of NaNO3 to CaCO3 = 2 : 1

Moles of CaCO3 = mass/molar mass

Mass of CaCO3 = 20 g; molar mass of CaCO3 = 100 g

Moles of CaCO3 = 20 g/100 g/mol = 0.2 moles

Moles of NaNO3 = 2 × 0.2 moles = 0.4 moles

1 Mole of NaNO3 = 6.02 × 10²³ formula units

0.4 moles of NaNO3 = 0.4 × 6.02 × 10²³ = 2.41 × 1023 formula units

2. Equation of reaction : 2 H2O ----> 2 H2 + O2

Mole ratio of oxygen to water = 1 : 2

At STP contains 6.02 × 10²³ molecules = 1 mole of water

6.58 × 10²⁴ molecules = 6.58 × 10²⁴ molecules × 1 mole of water/ 6.02 × 10²³ molecules = 10.93 moles of water

Moles of oxygen gas produced = 10.93÷2 = 5.465 moles of oxygen gas

At STP, 1 mole of oxygen gas = 22.4 L

5.465 moles of oxygen gas = 5.465 moles × 22.4 L/1 mole = 122 L

3.Equation of reaction: 6 K + N2 ----> 2 K3N

Mole ratio of Nitrogen gas and potassium = 6 : 1

Moles potassium = mass/ molar mass

Mass of potassium = 90.0 g, molar mass of potassium = 39.0 g/mol

Moles of potassium = 90.0 g / 39.0 g/mol = 2.3077moles

Moles of Nitrogen gas = 2.3077 moles / 6 = 0.3846 moles

At STP, 1 mole of nitrogen gas = 22.4 L

0.3486 moles of oxygen gas = 0.3486 moles × 22.4 L/1 mole = 7.81 L

shubhamchouhanvrVT

1. Volume of oxygen 2.41 × 1023   units

2. Nitrogen gas 122 L

3. 7.81 L

What will be the volume of oxygen and nitrogen?

Equation of the reaction:

[tex]\rm 2Na(No_3)+Ca(Co_3)---- > Na_2(Co_3)+Ca(No_3)_2[/tex]

[tex]NaNo_3\ and \ CaCo_3=\dfrac{2}{1}[/tex]

[tex]\rm Moles \ of \ CaCo_3=\dfrac{Mass}{Molar\ Mass}[/tex]

Mass of CaCO3 = 20 g; molar mass of CaCO3 = 100 g

[tex]\rm \ Moles \ of \ CaCo_3=\dfrac{20}{100} =0.2[/tex]

[tex]Moles \ of \ NaNo_3=2\times 0.2=0.4\ moles[/tex]

[tex]1\ mole\ of \ NaNo_3=6.02\times10^{23}[/tex]

0.4 moles of NaNO3 = [tex]0.4\times 6.02\times 10^{23}=2.41\times 10^{23}[/tex]  

2. Equation of reaction :[tex]2H_2o---- > 2H_2+o_2[/tex]  

The mole ratio of oxygen to water =  [tex]\dfrac{2}{1}[/tex]

At STP contains

 [tex]\dfrac{6.02\times 10^{24}\times 1 \ mole \ of \ water}{6.02\times 10^{23}} =10.93\ moles \ of \ water[/tex]

Moles of oxygen gas produced =

[tex]\dfrac{10.93}{2} =5.465\ Moles \ of \ oxygen \ gas[/tex]

At STP, 1 mole of oxygen gas = 22.4 L

5.465 moles of oxygen gas =    [tex]5.465\times 22.4=122\ L[/tex]  

3. Equation of reaction:

[tex]6k+N_2---- > 2K_3N[/tex]6

The mole ratio of Nitrogen gas and potassium =   [tex]\dfrac{6}{1}[/tex]

[tex]Moles\ of \ potassium = \dfrac{Mass}{Molar\ Mass}[/tex]

Mass of potassium = 90.0 g,

molar mass of potassium = 39.0 g/mol

[tex]Moles\ of \ potassium = \dfrac{90}{39}=2.3077\ moles[/tex]

[tex]Moles\ of \ Nitrogen = \dfrac{ 2.3077}{6}=0.3846\ moles[/tex]

At STP, 1 mole of nitrogen gas = 22.4 L

0.3486 moles of oxygen gas =

[tex]0.3486\times 22.4=7.81\ L[/tex]

Thus

1. Volume of oxygen 2.41 × 1023   units

2. Nitrogen gas 122 L

3. 7.81 L

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