H₂ + ½O₂ → H₂O
2g → 18g
So, 18g water needs 2g H₂
So, 1g water needs 2/18g H₂
So, 180g water needs 2/18 x 180g H₂
→ 20 litres of H₂
Answer:
0.00335 moles
Explanation:
From the question, Using
PV = nRT................... Equation 1
Where P = pressure, V = Volume, n = number of moles of argon gas, R = Molar gas constant, T = Temperature.
make n the subject of the equation
n = PV/RT............... Equation 2
Given: P = 1 atm (standard pressure), T = 273 K (standard temperature), V = 75 mL = 0.075 dm³
Constant: R = 0.082 atm·dm³/K·mol
Substitute into equation 2
n = (1×0.075)/(273×0.082)
n = 0.075/22.386
n = 0.00335 moles
Formula : BaI₂. <span>barium iodide</span>
Answer:
Option B. 4 moles of the gaseous product
Explanation:
Data obtained from the question include:
Initial volume (V1) = V
Initial number of mole (n1) = 2 moles
Final volume (V2) = 2V
Final number of mole (n2) =..?
Applying the Avogadro's law equation, we can obtain the number of mole of the gaseous product as follow:
V1/n1 = V2/n2
V/2 = 2V/n2
Cross multiply
V x n2 = 2 x 2V
Divide both side by V
n2 = (2 x 2V)/V
n2 = 2 x 2
n2 = 4 moles
Therefore, 4 moles of the gaseous product were produced.
