Answer:
V CH4(g) = 190.6 L
Explanation:
assuming ideal gas:
∴ STP: T =298 K and P = 1 atm
∴ R = 0.082 atm.L/K.mol
∴ moles (n) = 7.80 mol CH4(g)
∴ Volume CH4(g) = ?
⇒ V = RTn/P
⇒ V CH4(g) = ((0.082 atm.L/K.mol)×(298 K)×(7.80 mol)) / (1 atm)
⇒ V CH4(g) = 190.6 L
Yes. bromine and sodium iodide can react to form sodium bromine and free iodine
Answer:
Explanation:
This is an example of a limiting reactant question, and is very common as a general chemistry problem.
We first see the balanced equation, that is:
2CuCl2+4KI→2CuI+4KCl+I2
We first need to find the limiting reactant
We see that 0.56 g of copper(II) chloride (CuCl2) reacts with 0.64 g of potassium iodide (KI) . So, let's convert those amounts into moles.
Copper(II) chloride has a molar mass of
134.45 g/mol . So in 0.56 g of copper(II) chloride, then there exist
0.56g134.45g/mol≈4.17⋅10−3 mol
Potassium iodide has a molar mass of
166 g/mol . So, in 0.64 g of potassium iodide, there exist
if it wrong i am sorry
