Answer :
1)Mass of O₂ produced from 88.4 g KCIO₃: 34.56 g.
2)Moles of O₂ needed for 9.81 mol C₆H₁₂O₆: 58.86 mol.
3)Mass of O₂ needed to burn 51.9 g CH₄: 207.616 g.
4)Mass of NaI needed to produce 97.6 g I₂: 57.542 g.
5)Volume of Cu(NO₃)₂ solution cannot be determined with the given information.
To determine the quantities of substances involved in the given reactions, we'll use stoichiometry and the molar masses of the compounds.
For the first reaction: 2 KCIO₃(s) → 2 KCI(s) + 3 O₂(g)
From the equation, we can see that 2 moles of KCIO₃ yield 3 moles of O₂.
Given 88.4 g of KCIO₃, we need to convert it to moles using its molar mass:
Molar mass of KCIO₃ = (39.1 g/mol) + (35.5 g/mol) + (3 * 16.0 g/mol) = 122.6 g/mol
Moles of KCIO₃ = 88.4 g / 122.6 g/mol = 0.72 mol
Therefore, using the stoichiometry, 0.72 mol KCIO₃ will yield (3/2) * 0.72 = 1.08 mol O₂.
The molar mass of O₂ is 32.0 g/mol, so the mass of O₂ produced is:
Mass of O₂ = 1.08 mol * 32.0 g/mol = 34.56 g
For the second reaction: 6 CO₂(g) + 6 H₂O(g) → C₆H₁₂O₆(s) + 6 O₂(g)
From the equation, we can see that 6 moles of CO₂ react with 6 moles of O₂.
Given 9.81 mol C₆H₁₂O₆, we can determine the required moles of O₂:
Moles of O₂ = 6 * 9.81 mol = 58.86 mol
For the third reaction: CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g)
We can see that 1 mole of CH₄ requires 2 moles of O₂.
Given 51.9 g of CH₄, we need to convert it to moles using its molar mass:
Molar mass of CH₄ = 12.0 g/mol + 4 * (1.0 g/mol) = 16.0 g/mol
Moles of CH₄ = 51.9 g / 16.0 g/mol = 3.244 mol
Therefore, using the stoichiometry, 3.244 mol CH₄ will require 2 * 3.244 = 6.488 mol O₂.
The molar mass of O₂ is 32.0 g/mol, so the mass of O₂ needed is:
Mass of O₂ = 6.488 mol * 32.0 g/mol = 207.616 g
For the fourth reaction: 2 NaI(aq) + Cl₂(g) → 2 NaCl(aq) + I₂(s)
We are given the mass of iodine, and we need to find the mass of sodium iodide.
The molar mass of I₂ is 253.8 g/mol, so the moles of I₂ produced are:
Moles of I₂ = 97.6 g / 253.8 g/mol = 0.384 mol
According to the stoichiometry, 2 moles of NaI produce 1 mole of I₂.
Therefore, moles of NaI = 0.384 mol / 1 = 0.384 mol.
The molar mass of NaI is (22.99 g/mol) + (126.9 g/mol) = 149.89 g/mol.
The mass of NaI needed is: Mass of NaI = 0.384 mol * 149.89 g/mol = 57.542 g.
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The volume of the solution is 390.0 milliliters.
1. For the first question about the production of oxygen from heating potassium chlorate, the balanced equation is: 2 KCIO3(s) -> 2 KCl(s) + 3 O2(g). From the equation, we can see that 2 moles of KCIO3 will produce 3 moles of O2. To calculate the grams of O2 produced, we need to use the molar mass of O2, which is 32 g/mol.
To calculate the grams of O2 produced from 88.4 g of KCIO3, we need to use the molar mass of KCIO3, which is 122.55 g/mol.
First, we calculate the number of moles of KCIO3: 88.4 g KCIO3 / 122.55 g/mol = 0.721 mol KCIO3.
Since the mole ratio between KCIO3 and O2 is 2:3, we can calculate the moles of O2 produced: 0.721 mol KCIO3 * (3 mol O2 / 2 mol KCIO3) = 1.0815 mol O2.
Finally, we calculate the grams of O2 produced: 1.0815 mol O2 * 32 g/mol = 34.568 g O2.
Therefore, heating 88.4 g of KCIO3 will produce approximately 34.568 grams of O2.
2. For the second question about the production of glucose in plants through photosynthesis, the balanced equation is: 6 CO2 + 6 H2O -> C6H12O6 + 6 O2. From the equation, we can see that 6 moles of CO2 will produce 6 moles of O2.
To calculate the moles of CO2 needed to produce 9.81 mol of C6H12O6, we need to use the mole ratio between CO2 and C6H12O6, which is 6:1.
Therefore, the number of moles of CO2 needed is 9.81 mol C6H12O6 * (6 mol CO2 / 1 mol C6H12O6) = 58.86 mol CO2.
3. For the third question about the combustion of propane, the balanced equation is: C3H8 + 5 O2 -> 3 CO2 + 4 H2O. From the equation, we can see that 5 moles of O2 are needed to completely burn 1 mole of C3H8.
To calculate the grams of O2 needed to completely burn 51.9 g of C3H8, we need to use the molar mass of C3H8, which is 44.1 g/mol.
First, we calculate the number of moles of C3H8: 51.9 g C3H8 / 44.1 g/mol = 1.177 mol C3H8.
Since the mole ratio between C3H8 and O2 is 1:5, we can calculate the moles of O2 needed: 1.177 mol C3H8 * (5 mol O2 / 1 mol C3H8) = 5.885 mol O2.
Finally, we calculate the grams of O2 needed: 5.885 mol O2 * 32 g/mol = 188.32 g O2.
Therefore, 188.32 grams of O2 are needed to completely burn 51.9 grams of C3H8.
4. For the fourth question about the preparation of iodine, the balanced equation is: 2 NaI(aq) + Cl2(g) -> 2 NaCl(aq) + I2(s). From the equation, we can see that 2 moles of NaI will produce 1 mole of I2.
To calculate the grams of NaI needed to produce 97.6 g of I2, we need to use the molar mass of NaI, which is 149.89 g/mol.
First, we calculate the number of moles of I2: 97.6 g I2 / 253.81 g/mol = 0.384 mol I2.
Since the mole ratio between NaI and I2 is 2:1, we can calculate the moles of NaI needed: 0.384 mol I2 * (2 mol NaI / 1 mol I2) = 0.768 mol NaI.
Finally, we calculate the grams of NaI needed: 0.768 mol NaI * 149.89 g/mol = 115.75 g NaI.
Therefore, 115.75 grams of NaI must be used to produce 97.6 grams of I2.
5. For the fifth question about the volume of a solution, we are given that 8.04 g of Cu(NO3)2 is dissolved in water to make a 0.110 M solution. To calculate the volume of the solution in milliliters, we can use the formula:
Volume (in mL) = (mass (in g) / molar mass (in g/mol)) / concentration (in mol/L).
First, we calculate the number of moles of Cu(NO3)2: 8.04 g / 187.55 g/mol = 0.0429 mol.
Then, we can calculate the volume of the solution: [tex](0.0429 mol / 0.110 mol/L) * 1000 mL/L[/tex] = 390.0 mL.
Therefore, the volume of the solution is 390.0 milliliters.
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