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
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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
Hint, convert grams to mols
so what? you still won't know what you're doing...
I assume all problems are at STP where a mol of gas is 22.4 Liters
http://www.jiskha.com/display.cgi?id=1429736766
1. To find the number of liters of carbon dioxide produced when 90.0 g of glucose completely reacts with oxygen, you need to use the stoichiometry of the balanced chemical equation.
First, calculate the moles of glucose:
Molar mass of glucose (C6H12O6) = 180.16 g/mol
Moles of glucose = mass of glucose / molar mass = 90.0 g / 180.16 g/mol = 0.4997 mol (rounded to four decimal places)
From the balanced chemical equation, you can see that for every 1 mol of glucose (C6H12O6), 6 moles of carbon dioxide (CO2) are produced.
Moles of carbon dioxide = moles of glucose × (6 moles CO2 / 1 mol glucose) = 0.4997 mol × 6 = 2.9982 mol (rounded to four decimal places)
Now, to convert moles of carbon dioxide to liters, you need to use the ideal gas law at standard temperature and pressure (STP):
1 mole of any gas occupies 22.4 liters at STP.
Liters of carbon dioxide = moles of carbon dioxide × 22.4 L/mol = 2.9982 mol × 22.4 L/mol = 67.13 L (rounded to two decimal places)
Therefore, the number of liters of carbon dioxide produced when 90.0 g of glucose completely reacts with oxygen is approximately 67.1 L.
Answer: C. 67.1 L
2. To find the number of liters of chlorine that can be produced when 1.96 L of HCl react with excess O2 at STP, you can use the stoichiometry of the balanced chemical equation.
From the balanced chemical equation, you can see that for every 4 moles of hydrochloric acid (HCl), 2 moles of chlorine (Cl2) are produced.
Moles of HCl = volume of HCl (in liters) / molar volume at STP = 1.96 L / 22.4 L/mol = 0.0875 mol (rounded to four decimal places)
Since the ratio of moles of HCl to moles of chlorine is 4:2, you can determine the moles of chlorine produced:
Moles of chlorine = moles of HCl × (2 moles Cl2 / 4 moles HCl) = 0.0875 mol × 0.5 = 0.04375 mol (rounded to five decimal places)
Now, convert moles of chlorine to liters using the molar volume at STP:
Liters of chlorine = moles of chlorine × molar volume at STP = 0.04375 mol × 22.4 L/mol = 0.98 L (rounded to two decimal places)
Therefore, the number of liters of chlorine produced when 1.96 L of HCl reacts with excess O2 at STP is approximately 0.98 L.
Answer: B. 0.98 L
3. To find the mass of KClO3 needed to produce 10.0 L of O2 at STP, you will again use the stoichiometry of the balanced chemical equation.
From the balanced chemical equation, you can see that for every 3 moles of oxygen gas (O2), 2 moles of potassium chlorate (KClO3) are required.
Moles of oxygen gas = volume of oxygen gas (in liters) / molar volume at STP = 10.0 L / 22.4 L/mol = 0.4464 mol (rounded to four decimal places)
Since the ratio of moles of O2 to moles of KClO3 is 3:2, you can determine the moles of KClO3 needed:
Moles of KClO3 = moles of O2 × (2 moles KClO3 / 3 moles O2) = 0.4464 mol × (2/3) ≈ 0.2976 mol (rounded to four decimal places)
Now, calculate the mass of KClO3 needed using its molar mass:
Molar mass of KClO3 = 122.55 g/mol
Mass of KClO3 = moles of KClO3 × molar mass = 0.2976 mol × 122.55 g/mol ≈ 36.44 g (rounded to two decimal places)
Therefore, approximately 36.5 grams of KClO3 are needed to produce 10.0 L of O2 at STP.
Answer: C. 36.5 g