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
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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
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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
See my response above.
1. To find the number of liters of carbon dioxide produced when 90.0 g of glucose reacts, we need to use stoichiometry to convert from grams of glucose to liters of carbon dioxide.
First, we determine the molar mass of glucose (C6H12O6). Carbon (C) has a molar mass of 12.01 g/mol, hydrogen (H) has a molar mass of 1.01 g/mol, and oxygen (O) has a molar mass of 16.00 g/mol. Since there are 6 carbons, 12 hydrogens, and 6 oxygens in glucose, the molar mass of glucose is:
(6 * 12.01 g/mol) + (12 * 1.01 g/mol) + (6 * 16.00 g/mol) = 180.18 g/mol.
Now we calculate the number of moles of glucose by dividing the mass (90.0 g) by the molar mass (180.18 g/mol):
moles of glucose = 90.0 g / 180.18 g/mol = 0.499 mol.
According to the balanced equation, for every mole of glucose, 6 moles of carbon dioxide are produced. Therefore, we can calculate the number of moles of carbon dioxide produced by multiplying the moles of glucose by the stoichiometric ratio:
moles of carbon dioxide = 0.499 mol glucose * 6 mol CO2 / 1 mol glucose = 2.994 mol CO2.
To convert from moles to liters of carbon dioxide, we need to use the ideal gas law at standard temperature and pressure (STP). At STP, one mole of any ideal gas occupies 22.4 L. Therefore, we can convert moles of carbon dioxide to liters:
liters of carbon dioxide = 2.994 mol CO2 * 22.4 L/mol = 67.14 L.
Thus, the answer is C. 67.1L.
2. To find the number of liters of chlorine gas produced when 1.96 L of HCl reacts, we need to use stoichiometry to convert from liters of HCl to liters of chlorine gas.
Based on the balanced equation, the stoichiometric ratio between HCl and Cl2 is 4:2, or simply 2:1. Therefore, the number of liters of chlorine gas produced will be half of the number of liters of HCl used.
liters of chlorine gas = 1.96 L / 2 = 0.98 L.
Thus, the answer is B. 0.98 L.
3. To find the mass of KClO3 needed to produce 10.0 L of O2 at STP, we need to use stoichiometry and the ideal gas law.
According to the balanced equation, for every 3 moles of oxygen gas produced, 2 moles of KClO3 are consumed. Therefore, we need to calculate the number of moles of KClO3 required to produce 10.0 L of O2.
First, we convert liters of O2 to moles of O2 using the ideal gas law:
moles of O2 = 10.0 L O2 * (1 mol O2 / 22.4 L) = 0.446 mol O2.
Since the stoichiometric ratio between O2 and KClO3 is 3:2, the number of moles of KClO3 required will be:
moles of KClO3 = 0.446 mol O2 * (2 mol KClO3 / 3 mol O2) = 0.297 mol KClO3.
Finally, we calculate the mass of KClO3 by multiplying the moles of KClO3 by its molar mass. The molar mass of KClO3 is the sum of the molar masses of potassium (K), chlorine (Cl), and three oxygen atoms (O):
molar mass of KClO3 = (39.10 g/mol K) + (35.45 g/mol Cl) + (3 * 16.00 g/mol O) = 122.55 g/mol.
mass of KClO3 = 0.297 mol KClO3 * 122.55 g/mol KClO3 = 36.50 g.
Thus, the answer is C. 36.5g.