The oxidation of phosphine (PH3) to phosphorus pentoxide (P2O5) is given by the chemical equation:
_PH3+O2-->_P2O5+H2O
A.balance the chemical equation
B.how many grams of O2 will react completely with 35.0grams of PH3?
C.how many of P2O5 and H2O will be produced in the reaction?
D.if the reaction gives 64% yield,how many grams of PH3 should be used to produce an actual yield of 250grams of P2O5?
go thru this similar problem (answer is on second page). Then try this one, I will be happy to check your work. http://www.wou.edu/las/physci/poston/ch221/pdf/Limiting-Reagent-Problems-Key.pdf
Where can i get the answer? Help please
A. To balance the chemical equation, we need to make sure that there is an equal number of atoms on both sides of the equation. Let's start by looking at the number of phosphorus (P) atoms. We have one P on the left side and two P's on the right side. To balance the P, we can put a coefficient of 2 in front of PH3 and P2O5, so the equation becomes:
2PH3 + O2 --> P2O5 + H2O
Now let's balance the hydrogen (H) atoms. We have six H atoms on the right side (2 from H2O), so we need to put a coefficient of 3 in front of H2O:
2PH3 + O2 --> P2O5 + 3H2O
Finally, let's balance the oxygen (O) atoms. We have four O atoms on the left side (2 from O2), so we need to put a coefficient of 2 in front of O2:
2PH3 + 2O2 --> P2O5 + 3H2O
B. To determine how many grams of O2 are needed to react completely with 35.0 grams of PH3, we first need to calculate the molar mass of PH3 and O2. The molar mass of PH3 is approximately 33.97 g/mol (1 phosphorus + 3 hydrogen), and the molar mass of O2 is approximately 32.00 g/mol (16.00 g/mol per oxygen atom).
Next, we need to convert the grams of PH3 to moles using the molar mass of PH3:
35.0 g PH3 * (1 mol PH3/33.97 g PH3) = 1.03 mol PH3
According to the balanced equation, we need 2 moles of O2 to react with 2 moles of PH3. Therefore, the number of moles of O2 needed is also 1.03 mol.
Finally, let's convert the moles of O2 to grams using the molar mass of O2:
1.03 mol O2 * (32.00 g O2/1 mol O2) = 32.96 g O2
Therefore, 32.96 grams of O2 will react completely with 35.0 grams of PH3.
C. The balanced equation shows that 2 moles of P2O5 and 3 moles of H2O are produced for every 2 moles of PH3 that react. Thus, the molar ratio of P2O5 to PH3 is 2:1, and the molar ratio of H2O to PH3 is 3:2.
To find the amount of P2O5 and H2O produced, we can use the calculated moles of PH3 from part B (1.03 mol).
For P2O5:
1.03 mol PH3 * (2 mol P2O5/2 mol PH3) = 1.03 mol P2O5
For H2O:
1.03 mol PH3 * (3 mol H2O/2 mol PH3) = 1.55 mol H2O
Therefore, in the reaction, 1.03 moles of P2O5 and 1.55 moles of H2O will be produced.
D. The given yield is 64%, which means that only 64% of the theorethical yield is obtained. We need to calculate the theorethical yield using the stoichiometry of the balanced equation.
Using the balanced equation, we know that for every 2 moles of PH3, 2 moles of P2O5 are produced. Therefore, the theorethical yield of P2O5 can be calculated as follows:
Theorethical yield of P2O5 = (250 g P2O5) * (1 mol P2O5/141.95 g P2O5) * (2 mol PH3 / 2 mol P2O5) * (97.97 g PH3 / 1 mol PH3)
Simplifying this calculation, we can find the theorethical yield of PH3:
Theorethical yield of P2O5 = (250 g P2O5) * (97.97 g PH3/141.95 g P2O5) = 173.5 g PH3
Now we can find out how much PH3 should be used to produce the given actual yield of 250 g P2O5 with 64% yield:
Actual Yield = (64% of Theorethical yield) = (64% * 173.5 g PH3) = 111.04 g PH3
Therefore, to produce an actual yield of 250 grams of P2O5 with a 64% yield, approximately 111.04 grams of PH3 should be used.