C the products are on the right side of the arrow
Answer:
Explanation:
2KClO₃ = 2KCl + 3O₂
2 moles of potassium chlorate gives 3 moles of oxygen gas
20 moles of potassium chlorate will give 30 moles of oxygen gas .
30 moles of oxygen = 30 x 6.02 x 10²³ molecules of oxygen .
= 180.6 x 10²³ molecules of O₂ .
Answer: The ΔH of the reaction if 51.3 g of 
reacts with excess 
to yield 1387.6 kJ is 432.27kJ
Explanation:
To calculate the moles, we use the equation:
moles of
As is present in excess, is the limiting reagent as it limits the formation of product.
If 3.21 moles of methane releases heat = 1387.6 kJ
Thus 1 mole of methane release=
Answer:
0.337 dm³
Explanation:
M(K2CO3) = 138.2 g/mol
1) 11.7 g * (1 mol/138.2 g) =0.08466 mol
2) 0.08466 mol * (1 dm³/0.25 mol) = 0.3386 dm³ ≈ 0.337 dm³
Answer:
Explanation:
We can use the Ideal Gas Law and solve for T.
pV = nRT
Data
p = 2.91 atm
V = 78.13 L
n = 3.07 mol
R = 0.082 06 L·atm·K⁻¹mol⁻¹
Calculations
