Final answer:
Using the balanced equation and the ideal gas law, we can determine the number of moles of CO2 produced. From there, we can use the mole ratio to find the number of moles of water produced and convert it to grams. The mass of H₂O produced in this reaction is 9.95g. The closest option is B. 10.8.
Step-by-step explanation:
Based on the balanced equation for the complete combustion of a hydrocarbon, we can calculate the mass of H2O produced using the information provided. From the equation, we know that for every 1 mole of the hydrocarbon, 8 moles of water (H2O) are produced. We are given that 7.2 g of the hydrocarbon produces 11.2 L of CO2 at STP (Standard Temperature and Pressure). To find the mass of water produced, we can use the molar mass of CO2 and the ideal gas law to calculate the number of moles of CO2 produced. Then, we can use the mole ratio from the balanced equation to find the number of moles of water produced and finally convert it to grams.
First, calculate the number of moles of CO2 produced using the ideal gas law:
PV = nRT
where P = pressure, V = volume, n = number of moles, R = ideal gas constant, and T = temperature
At STP (Standard Temperature and Pressure),
P = 1 atm
V = 11.2 L
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T = 273.15 K (standard temperature)
Substituting the values:
1 atm * 11.2 L = n * 0.0821 L·atm/(mol·K) * 273.15 K
n = 0.484 mol CO2
Using the mole ratio from the balanced equation, we know that for every 7 moles of CO2 produced, 8 moles of water are produced:
n_water = (0.484 mol CO2 * 8 mol water) / 7 mol CO2
n_water = 0.552 mol water
Finally, convert the number of moles of water to grams using the molar mass of water:
m_water = 0.552 mol * (18.02 g/mol water) = 9.95 g
Therefore, the mass of water produced in this reaction is 9.95 g.