Given the equation C3H8+5O2->3CO2+4H2O.
1.)Suppose 1.5 moles of C3H8 will be used, How many moles of H2O and CO2 will be formed?
2.)what volumes (at STP) of O2 and CO2 are involved?
1.) Well, let's do some mole math! According to the balanced equation, for every 1 mole of C3H8, 3 moles of CO2 and 4 moles of H2O are formed. Since we have 1.5 moles of C3H8, we can double those numbers to get our answer. Therefore, we'll have 3 moles of CO2 and 4 moles of H2O formed.
2.) Ah, volumes at STP, it's like a chemistry vacation! Okay, so we need to know the molar volume at STP, which is 22.4 liters per mole. Now, let's look at our equation again. For every 3 moles of CO2, we also have 5 moles of O2 reacting. So, for our 1.5 moles of CO2, we'll have 2.5 moles of O2.
Using the molar volume at STP, we can convert those moles into volumes. Therefore, we'll have 33.6 liters of O2 and 33.6 liters of CO2 involved in this hilarious experiment. Happy chemistry-ing!
To determine the number of moles of CO2 and H2O formed, we need to use the stoichiometric coefficients from the balanced equation.
1.) For every 1 mole of C3H8 (propane), we get 3 moles of CO2 and 4 moles of H2O. Since we are using 1.5 moles of C3H8, we can calculate the number of moles of CO2 and H2O formed:
CO2: (1.5 moles C3H8) x (3 moles CO2/1 mole C3H8) = 4.5 moles CO2
H2O: (1.5 moles C3H8) x (4 moles H2O/1 mole C3H8) = 6 moles H2O
Therefore, 4.5 moles of CO2 and 6 moles of H2O will be formed.
2.) To determine the volumes of O2 and CO2 involved at STP (Standard Temperature and Pressure), we will use the ideal gas law. At STP, 1 mole of any ideal gas occupies 22.4 liters.
From the equation, we know that the stoichiometric ratio of O2 to CO2 is 5:3. Therefore, for every 5 moles of O2, we get 3 moles of CO2.
O2: (1.5 moles C3H8) x (5 moles O2/1 mole C3H8) = 7.5 moles O2
CO2: (1.5 moles C3H8) x (3 moles CO2/1 mole C3H8) = 4.5 moles CO2
At STP, 1 mole of gas occupies 22.4 liters. Therefore:
Volume of O2 = (7.5 moles O2) x (22.4 liters/mole) = 168 liters
Volume of CO2 = (4.5 moles CO2) x (22.4 liters/mole) = 100.8 liters
Therefore, 168 liters of O2 and 100.8 liters of CO2 are involved at STP.
To answer these questions, we need to use stoichiometry, which relates the amounts of substances involved in a chemical reaction. The coefficients in the balanced equation give us the stoichiometric ratios.
1.) To find the number of moles of CO2 and H2O formed, we first need to calculate the limiting reagent. In this case, C3H8 is our limiting reagent because we have the given quantity of 1.5 moles.
The stoichiometric ratio between C3H8 and CO2 is 1:3, which means for every 1 mole of C3H8, 3 moles of CO2 are produced. Therefore, 1.5 moles of C3H8 will produce (3 moles CO2/1 mole C3H8) * 1.5 moles C3H8 = 4.5 moles of CO2.
Similarly, the stoichiometric ratio between C3H8 and H2O is 1:4, which means for every 1 mole of C3H8, 4 moles of H2O are produced. Therefore, 1.5 moles of C3H8 will produce (4 moles H2O/1 mole C3H8) * 1.5 moles C3H8 = 6 moles of H2O.
Therefore, 1.5 moles of C3H8 will produce 4.5 moles of CO2 and 6 moles of H2O.
2.) To find the volumes of O2 and CO2 at Standard Temperature and Pressure (STP), we need to use the ideal gas law. At STP, 1 mole of gas occupies approximately 22.4 liters.
The stoichiometric ratio between C3H8 and O2 is 1:5, which means for every 1 mole of C3H8, 5 moles of O2 are required. Therefore, 1.5 moles of C3H8 will require (5 moles O2/1 mole C3H8) * 1.5 moles C3H8 = 7.5 moles of O2.
Using the ideal gas law, we can calculate the volume of O2 at STP:
V(O2) = n(O2) * 22.4 L/mol = 7.5 moles * 22.4 L/mol = 168 L
Similarly, the stoichiometric ratio between CO2 and C3H8 is 3:1, which means for every 3 moles of CO2 produced, 1 mole of C3H8 is consumed. Therefore, 1.5 moles of C3H8 will produce (3 moles CO2/1 mole C3H8) * 1.5 moles C3H8 = 4.5 moles of CO2.
Using the ideal gas law, we can calculate the volume of CO2 at STP:
V(CO2) = n(CO2) * 22.4 L/mol = 4.5 moles * 22.4 L/mol = 100.8 L
Therefore, approximately 168 liters of O2 and 100.8 liters of CO2 will be involved in the reaction at STP.
1. The coefficients tell you that
1 mol C3H8 produces 3 mol CO2 and 4 mols H2O. So 1.5 mol C3H8 will produce ...mols CO2 and ....mols H2O?
2. 1 mol of a gas at STP occupies 22.4 L.
So your mols CO2 and your mols H2O will occupy ...... L each.