1.Cell respiration glucose is reacted wth oxygen in body to produce carbon dioxide and water how many liters of carbon dioxide would be produced if 90.0 g of glucose completely reacts with oxygen
C6H12O6(s)+ 6O2(g) -> 6H2O(g) + 6CO2 (g)
How many liters of carbon dioxide would be produced if 90.0 g of glucose completely reacts with oxygen
Show Work
A.11.2L B.21.99L. C.67.1L. D.131.9L.
2. How many liters of chlorine as can be produced when 1.96 L of HCl react with excess O2 at STP
4HCl(g)+O2(g)->2 Cl(g)+ 2H2O(l)
A.0.49 B. 0.98L. C. 1.96 L D. 3.92 L
Show work
3.The decomposition of potassium chlorate gives oxygen gas according to the reaction:
2 KClO3(s) -> 2KCl(s)+3O2(g)
How many grams KClO3 are needed to produce 10.0 L of O2 at STP
A. 18.2g. B.25.5g. C.36.5g. D.54.7g
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1. To calculate the number of liters of carbon dioxide produced when 90.0 g of glucose completely reacts with oxygen, we need to use stoichiometry, which involves converting the mass of one substance to the number of moles, and then using the balanced chemical equation to determine the corresponding number of moles of another substance.
First, we need to convert the mass of glucose (C6H12O6) to moles. The molar mass of glucose is 180.16 g/mol.
Number of moles of glucose = mass of glucose / molar mass of glucose
Number of moles of glucose = 90.0 g / 180.16 g/mol
Number of moles of glucose = 0.4998 mol (approximately 0.5 mol)
According to the balanced chemical equation, we know that 1 mol of glucose reacts to produce 6 mol of carbon dioxide.
Now, we can use this ratio to find the number of moles of carbon dioxide produced:
Number of moles of carbon dioxide = number of moles of glucose × (6 mol CO2 / 1 mol glucose)
Number of moles of carbon dioxide = 0.5 mol × 6
Number of moles of carbon dioxide = 3 mol
To convert the number of moles of carbon dioxide to volume at STP, we need to use the ideal gas law. At STP, 1 mole of any gas occupies 22.4 liters.
Number of liters of carbon dioxide = number of moles of carbon dioxide × 22.4 L/mol
Number of liters of carbon dioxide = 3 mol × 22.4 L/mol
Number of liters of carbon dioxide = 67.2 L (approximately 67.1 L)
Therefore, the correct answer is option C. 67.1 L.
2. To calculate the number of liters of chlorine gas produced when 1.96 L of HCl reacts with excess oxygen at STP, we need to use the balanced chemical equation and stoichiometry.
According to the balanced chemical equation, 4 moles of HCl react to produce 2 moles of Cl2 gas.
First, let's convert the volume of HCl to moles using the ideal gas law at STP.
Number of moles of HCl = volume of HCl / molar volume at STP
Number of moles of HCl = 1.96 L / 22.4 L/mol
Number of moles of HCl = 0.0875 mol (approximately 0.09 mol)
Now, we can use this ratio to find the number of moles of Cl2 gas produced:
Number of moles of Cl2 = number of moles of HCl × (2 mol Cl2 / 4 mol HCl)
Number of moles of Cl2 = 0.09 mol × 0.5
Number of moles of Cl2 = 0.045 mol
To convert the number of moles of Cl2 to volume at STP, we again use the ideal gas law:
Number of liters of Cl2 = number of moles of Cl2 × 22.4 L/mol
Number of liters of Cl2 = 0.045 mol × 22.4 L/mol
Number of liters of Cl2 = 1.008 L (approximately 1.0 L)
Therefore, the correct answer is option C. 1.96 L.
3. To calculate the mass of potassium chlorate (KClO3) needed to produce 10.0 L of oxygen gas at STP, we will again use the balanced chemical equation and stoichiometry.
According to the balanced chemical equation, 2 moles of KClO3 react to produce 3 moles of O2 gas.
First, let's convert the volume of oxygen gas to moles using the ideal gas law at STP:
Number of moles of O2 = volume of O2 / molar volume at STP
Number of moles of O2 = 10.0 L / 22.4 L/mol
Number of moles of O2 = 0.4464 mol (approximately 0.45 mol)
Now, we can use this ratio to find the number of moles of KClO3:
Number of moles of KClO3 = number of moles of O2 × (2 mol KClO3 / 3 mol O2)
Number of moles of KClO3 = 0.45 mol × (2/3)
Number of moles of KClO3 = 0.3 mol
To calculate the mass of KClO3 needed, we need to multiply the number of moles by the molar mass of KClO3, which is 122.55 g/mol.
Mass of KClO3 = number of moles of KClO3 × molar mass of KClO3
Mass of KClO3 = 0.3 mol × 122.55 g/mol
Mass of KClO3 = 36.765 g (approximately 36.5 g)
Therefore, the correct answer is option C. 36.5 g.