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Answer:
76.0%
Explanation:
Let's consider the following reaction.
CaCO₃(s) ⇄ CaO(s) + CO₂(g)
At equilibrium, the equilibrium constant Kp is:
Kp = 1.16 = pCO₂ ⇒ pCO₂ = 1.16 atm
We can calculate the moles of CO₂ at equilibrium using the ideal gas equation.
From the balanced equation, we know that 1 mole of CO₂ is produced by 1 mole of CaCO₃. Taking into account that the molar mass of CaCO₃ is 100.09 g/mol, the mass of CaCO₃ that reacted is:
The percentage by mass of the CaCO₃ that reacted to reach equilibrium is:
Answer:
50.8 g
Explanation:
Equation of reaction.
From the given information, the number of moles of methane = mass/ molar mass
= 15.4 g / 16.04 g/mol
= 0.960 mol
number of moles of oxygen gas = 90.3 g / 32 g/ mol
= 2.82 mol
Since 1 mol of methane requires 2 moles of oxygen
Then 0.960 mol of methane will require = 0.960 mol × 2 = 1.92 mol of oxygen gas
Thus, methane serves as a limiting reagent.
2.82 mol oxygen gas will result in 2.82 moles of water
So, the theoretical yield of water = moles × molar mass
= 2.82 mol × 18.01528 g/mol
= 50.8 g