I just need help with these three
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)
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
Dr Bob told you much more precisely than I how do do this sort of problem. I suggest just doing them.
by the way, your book or Google will tell you how many grams of any element are in a mol.
for example for H about 1
for O about 16
for C about 12 etc
Do you not have a periodic table?
Please just do the problems.
1. To find the number of liters of carbon dioxide produced, we need to use the stoichiometry of the balanced equation.
Step 1: Calculate the molar mass of glucose (C6H12O6):
C: 6 * 12.01 g/mol = 72.06 g/mol
H: 12 * 1.01 g/mol = 12.12 g/mol
O: 6 * 16.00 g/mol = 96.00 g/mol
Total: 72.06 + 12.12 + 96.00 = 180.18 g/mol
Step 2: Calculate the number of moles of glucose:
90.0 g / 180.18 g/mol = 0.4996 mol
Step 3: Use the stoichiometry of the balanced equation to find the number of moles of carbon dioxide produced:
1 mol glucose reacts to produce 6 mol carbon dioxide
0.4996 mol glucose * 6 mol CO2 / 1 mol glucose = 2.9976 mol CO2
Step 4: Convert moles of carbon dioxide to liters at STP (Standard Temperature and Pressure):
1 mol of any gas occupies 22.4 L at STP
2.9976 mol CO2 * 22.4 L/mol = 67.144 L
Therefore, approximately 67.1 liters of carbon dioxide would be produced when 90.0 g of glucose completely reacts with oxygen.
2. To find the number of liters of chlorine gas produced, we need to use the stoichiometry of the balanced equation.
Step 1: Use the stoichiometry of the balanced equation:
4 mol HCl reacts to produce 2 mol Cl2
Step 2: Convert 1.96 L of HCl to moles using the ideal gas law at STP:
1 mol gas occupies 22.4 L at STP
1.96 L HCl * (1 mol HCl / 22.4 L) = 0.0875 mol HCl
Step 3: Use the stoichiometry of the balanced equation to find the number of moles of Cl2 produced:
0.0875 mol HCl * (2 mol Cl2 / 4 mol HCl) = 0.04375 mol Cl2
Step 4: Convert moles of chlorine gas to liters at STP:
0.04375 mol Cl2 * 22.4 L/mol = 0.98 L
Therefore, the answer is B. 0.98 L of chlorine gas can be produced when 1.96 L of HCl react with excess O2 at STP.
3. To find the grams of KClO3 needed to produce 10.0 L of O2 at STP, we need to use the stoichiometry of the balanced equation.
Step 1: Use the stoichiometry of the balanced equation:
2 mol KClO3 produces 3 mol O2
Step 2: Convert 10.0 L of O2 to moles using the ideal gas law at STP:
1 mol gas occupies 22.4 L at STP
10.0 L O2 * (1 mol O2 / 22.4 L) = 0.4464 mol O2
Step 3: Use the stoichiometry of the balanced equation to find the number of moles of KClO3 needed:
0.4464 mol O2 * (2 mol KClO3 / 3 mol O2) = 0.2976 mol KClO3
Step 4: Calculate the mass of KClO3 needed:
0.2976 mol KClO3 * molar mass of KClO3 = grams of KClO3
The molar mass of KClO3 is:
K: 1 * 39.10 g/mol = 39.10 g/mol
Cl: 3 * 35.45 g/mol = 106.35 g/mol
O: 3 * 16.00 g/mol = 48.00 g/mol
Total: 39.10 + 106.35 + 48.00 = 193.45 g/mol
0.2976 mol KClO3 * 193.45 g/mol = 57.59 g
Therefore, approximately 57.6 grams of KClO3 are needed to produce 10.0 L of O2 at STP. The answer is D. 54.7g.