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
John, you work all of these stoichiometry problems with 3 or 4 steps.
1. Write and balance the equation.
2. Convert what you have to mols. There are a couple of ways to do this.
a. If you have grams, then mols = grams/molar mass.
b. If you have a solution, then mols = M x L = ?
3. Use the coefficients in the balanced equation to convert mols of what you have to mols of what you want.
4. Then you convert mols of what you want (the product in most cases) to either (usually)
a. grams or
b. volume
If grams, then grams = mols x molar mass
If volume, then mols x 22.4L/mol if at STP or use PV = nRT if other than standard conditions are given.
With those steps you can work about 98% of the stoichiometry problems but I'm not about to do your test for you. That is cheating and you can't go through life cheating although some get by with it longer than others.
1. To solve this problem, we can use stoichiometry and the molar ratio between glucose and carbon dioxide in the balanced equation.
First, calculate the number of moles of glucose (C6H12O6) using its molar mass:
Molar mass of C6H12O6 = 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol
Number of moles of glucose = 90.0 g / 180.18 g/mol = 0.4997 mol (approximately)
Next, use the molar ratio between glucose and carbon dioxide from the balanced equation:
From the equation: 1 mol of glucose produces 6 mol of CO2
Number of moles of CO2 produced = 0.4997 mol × 6 = 2.997 mol (approximately)
Finally, convert the number of moles of CO2 to liters at STP:
1 mol of any ideal gas at STP occupies 22.4 L
Volume of CO2 produced = 2.997 mol × 22.4 L/mol = 67.1 L
Therefore, the answer is option C: 67.1 L.
2. For this problem, we can again use stoichiometry and the molar ratio between HCl and Cl2 in the balanced equation.
We start by converting the given volume of HCl from liters to moles at STP:
Volume of HCl = 1.96 L
Based on the ideal gas law, 1 mole of any ideal gas occupies 22.4 L at STP.
Number of moles of HCl = 1.96 L / 22.4 L/mol = 0.0875 mol (approximately)
Next, use the molar ratio between HCl and Cl2 from the balanced equation:
From the equation: 4 mol of HCl produces 2 mol of Cl2
Number of moles of Cl2 produced = 0.0875 mol × 2 / 4 = 0.04375 mol (approximately)
Finally, convert the number of moles of Cl2 to liters at STP:
Volume of Cl2 produced = 0.04375 mol × 22.4 L/mol = 0.98 L
Therefore, the answer is option B: 0.98 L.
3. This problem requires us to determine the number of grams of KClO3 needed to produce a given volume of O2 at STP.
First, convert the given volume of O2 from liters to moles at STP:
Volume of O2 = 10.0 L
Based on the ideal gas law, 1 mole of any ideal gas occupies 22.4 L at STP.
Number of moles of O2 = 10.0 L / 22.4 L/mol = 0.4464 mol (approximately)
Next, use the molar ratio between KClO3 and O2 from the balanced equation:
From the equation: 2 mol of KClO3 produces 3 mol of O2
Number of moles of KClO3 needed = 0.4464 mol × 2 / 3 = 0.2976 mol (approximately)
Finally, calculate the mass of KClO3 using its molar mass:
Molar mass of KClO3 = 39.10 + 35.45 + 3(16.00) = 122.55 g/mol
Mass of KClO3 needed = 0.2976 mol × 122.55 g/mol = 36.47 g (approximately)
Therefore, the answer is option C: 36.5 g.