<span>aluminium phosphide AlP</span>
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>Galvanized steel is preferred for outdoor uses because it is ideal to prevent rotting/corrosion
A steel will rot more quickly if it's exposed to a larger amount of oxygen and H2O , which will exist if we put it oudoor
Coating the steel with additional zinc will slow down the process</span>
Answer:
Explanation:
Two moles of magnesium (Mg) and five moles of oxygen (O2) are placed in a reaction vessel. When magnesium is ignited, it reacts with oxygen. What is the limiting reactant in this experiment?
Mg + O2 → MgO (unbalanced)
first, balance the equation
2Mg +O2-------> 2MgO
two magnesium atoms react with one diatomic oxygen molecule
there is a 1:1 ratio of magnesium to oxygen atoms
but we have 2 moles of magnesium atoms and 2X5 = 10 moles of oxygen atoms
the lesser magnesium LIMITS the amount of product we can make, so it is the LIMITING REAGENT.
