how many grams of oxygen gas are needed to react with 5.7 moles of phosphorus
P4 + 5O2 ==> P4O10
Using the coefficients in the balanced equation, convert moles P to moles oxygen.
Now convert moles oxygen to grams. g = moles x molar mass.
To answer this question, we need to first determine the balanced chemical equation for the reaction between oxygen gas (O2) and phosphorus (P). The balanced equation for the reaction is:
4P + 5O2 → 2P2O5
According to the balanced equation, 4 moles of phosphorus react with 5 moles of oxygen gas to produce 2 moles of phosphorus pentoxide (P2O5).
Now, we can use the stoichiometry of the balanced equation to calculate the number of moles of oxygen gas required to react with 5.7 moles of phosphorus.
Using the ratio of 5 moles of oxygen gas to 4 moles of phosphorus, we can set up the following proportion:
5 moles O2 / 4 moles P = X moles O2 / 5.7 moles P
To solve for X (the number of moles of oxygen gas), we can use cross multiplication:
5 moles O2 * 5.7 moles P = 4 moles P * X moles O2
28.5 moles O2 = 4X moles O2
Now, we can solve for X:
X = 28.5 moles O2 / 4
X = 7.125 moles O2
Therefore, 7.125 moles of oxygen gas are needed to react with 5.7 moles of phosphorus.
To convert moles of oxygen gas to grams, we need to use the molar mass of oxygen gas, which is 32 g/mol. Multiplying the number of moles (7.125) by the molar mass (32 g/mol) gives us the answer:
7.125 moles O2 * 32 g/mol = 228 grams of oxygen gas
Therefore, approximately 228 grams of oxygen gas are needed to react with 5.7 moles of phosphorus.