Answer :
The volume of O2 needed to react completely with 39.4 g of MoS2 at STP is approximately 18.4 liters.
To determine the volume of oxygen (O2) required to react completely with 39.4 g of MoS2 at STP (Standard Temperature and Pressure), we need to use stoichiometry. The balanced chemical equation for the reaction between MoS2 and O2 is 2 MoS2 + 7 O2 → 2 MoO3 + 4 SO2. From the equation, we can determine the molar ratio between MoS2 and O2. By converting the given mass of MoS2 to moles, we can then calculate the moles of O2 needed. Finally, using the ideal gas law and STP conditions, we can find the volume of O2.
The balanced chemical equation for the reaction between MoS2 and O2 is 2 MoS2 + 7 O2 → 2 MoO3 + 4 SO2. From the equation, we can determine the molar ratio between MoS2 and O2, which is 2:7. This means that for every 2 moles of MoS2, we need 7 moles of O2 for complete reaction.
First, we calculate the number of moles of MoS2 using its molar mass. The molar mass of MoS2 is given as 160.06 g/mol. Therefore, the number of moles of MoS2 is:
moles of MoS2 = mass of MoS2 / molar mass of MoS2
= 39.4 g / 160.06 g/mol
≈ 0.246 mol
Next, we use the molar ratio to determine the moles of O2 required:
moles of O2 = (moles of MoS2) × (7 moles of O2 / 2 moles of MoS2)
≈ 0.246 mol × (7/2)
≈ 0.861 mol
Finally, we can calculate the volume of O2 at STP using the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
At STP, the pressure (P) is 1 atmosphere (atm) and the temperature (T) is 273.15 Kelvin (K). The ideal gas constant (R) is approximately 0.0821 L·atm/(mol·K). Plugging in the values, we have:
V = (nRT) / P
= (0.861 mol) × (0.0821 L·atm/(mol·K)) × (273.15 K) / 1 atm
≈ 18.4 L
Learn more about stoichiometry here:
brainly.com/question/28780091
#SPJ11
