Answer:
We'll have 1 mol Al2O3 and 3 moles H2
Explanation:
Step 1: data given
Numer of moles of aluminium = 2 moles
Number of moles of H2O = 6 moles
Step 2: The balanced equation
2Al + 3H2O → Al2O3 + 3H2
Step 3: Calculate the limiting reactant
For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2
Aluminium is the limiting reactant. It will completely be consumed (2 moles).
H2O is in excess. There will react 3/2 * 2 = 3 moles
There will remain 6 - 3 = 3 moles
Step 4: Calculate moles products
For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2
For 2 moles Al we'll have 2/1 = 1 mol Al2O3
For 2 moles Al We'll have 3/2 * 2 = 3 moles H2
We'll have 1 mol Al2O3 and 3 moles H2
Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
<span>Heat from the can is making the water molecule move faster, which melts the ice.</span>
NaBrO3 is the chemical formula for Sodium Bromate.
