Oxygen gas and aqueous benzoic acid react to form carbon dioxide gas and water gas. Determine the mass of water in grams if 5.43×104 milligrams of oxygen reacts with an excess of benzoic acid.

I balanced the equation to
2C7H6O2+15O2=14CO2+6H2O

Convert milligrams of oxygen to grams, then divide by f.wt O2 (=32) => moles O2. Then apply equation ratios...

15 moles O2 => 6 moles HOH
N-moles O2 => x

(15/N-molesO2) = (6/X)
Solve for X and then multiply by f.wt. HOH (=18g/mole) => grams HOH.

To determine the mass of water produced in the reaction, we can use stoichiometry. From the balanced equation, we can see that 6 moles of water (H2O) are produced for every 2 moles of benzoic acid (C7H6O2) consumed.

First, we need to convert the given mass of oxygen from milligrams to grams:
5.43×104 milligrams = 5.43×104/1000 = 54.3 grams

Next, we can use the molar ratio of benzoic acid to water to find the number of moles of benzoic acid:
2 moles of benzoic acid (C7H6O2) = 6 moles of water (H2O)
Since we have an excess of benzoic acid, we can assume that all the oxygen reacts, thus we have:
54.3 grams of oxygen = moles of oxygen

To find the moles of benzoic acid:
moles of benzoic acid = (moles of oxygen) * (2 moles of benzoic acid / 15 moles of oxygen)
moles of benzoic acid = (54.3 / 32) × (2 / 15) = 0.227 moles

Finally, we can use the molar mass of water to calculate the mass of water produced:
molar mass of water (H2O) = (1 × 2) + 16 = 18 grams/mol
mass of water = moles of water × molar mass of water
mass of water = 0.227 × 18 = 4.086 grams

Therefore, the mass of water produced in the reaction is 4.086 grams.

To determine the mass of water produced in the reaction, we need to first calculate the number of moles of oxygen consumed in the reaction, and then use the stoichiometry of the balanced equation to find the number of moles of water produced. Finally, we can convert the number of moles of water to grams using the molar mass of water.

1. Convert the given mass of oxygen from milligrams to grams:
5.43 × 10^4 mg = 5.43 × 10^4/1000 g = 54.3 g

2. Calculate the number of moles of oxygen:
Moles = mass (g) / molar mass (g/mol)
The molar mass of oxygen (O2) = 32 g/mol
Moles of oxygen = 54.3 g / 32 g/mol = 1.6975 mol (rounded to 4 decimal places)

3. Use the balanced equation to find the mole ratio between oxygen and water:
From the balanced equation, the ratio of oxygen to water is 15:6
Therefore, Moles of water = (1.6975 mol) * (6/15) = 0.67875 mol (rounded to 5 decimal places)

4. Convert the number of moles of water to grams:
The molar mass of water (H2O) = 18 g/mol
Mass of water = Moles of water * molar mass of water
= 0.67875 mol * 18 g/mol
= 12.2175 g (rounded to 4 decimal places)

Therefore, the mass of water produced in the reaction is approximately 12.2175 grams.