B. Newton
Is the metric unit of force
Answer:
The limiting reactant is propene, 
.
Explanation:
Moles of nitrogen propene = 2 mol
Moles of oxygen = 10 mol
According to reaction, 2 moles of propene reacts with 9 moles of oxygen gas, then 2 moles of propene will react with:
According to the question, we have 10 moles of oxygen gas, which is more than 9 moles of oxygen gas. This indicates that propene is present in a limiting amount hence, it is a limiting reactant.
The limiting reactant is propene, hence the correct answer is the .
First we calculate the number of moles of sugar (which I assume is sucrose).
number of moles = mass / molecular weight
number of moles of sugar = 19 / 342 = 0.055 moles
Now we may calculate the molarity of the solution.
molarity = number of moles / solution volume (L)
molarity = 0.055 / 0.05 = 1.1 M
Answer;
12 carbon atoms, 22 hydrogen atoms and 11 oxygen atoms.
Explanation:
Using a balanced chemical equation we can identify the number of carbon, hydrogen, and oxygen atoms in sugar.
CxHyOn + 12O₂ → 11 H₂O + 12CO₂
When an equation is completely balanced, then the number of each atom of an element is equal on the reactant side and the product side.
Therefore;
For carbon; x = 12
For Hydrogen; y = (11×2) = 22
For Oxygen; n + (12×2) = 11 + (12×2)
= n + 24 = 11 + 24
n = 11
Therefore the sugar has, 12 carbon atoms, 22 hydrogen atoms and 11 oxygen atoms.
Thus the balanced equation would be;
C₁₂H₂₂O₁₁ + 12O₂ → 11 H₂O + 12CO₂
Answer:
1.16L
Explanation:
First, let us generate the balanced equation for the reaction. This is illustrated below
2Mg + O2 —> 2MgO
Now let us covert 2.5g of Mg given in the question to moles. This is illustrated below:
Molar Mass of Mg = 12g/mol
Mass of Mg = 2.5g
Number of mole of Mg =?
Number of mole = Mass /Molar Mass
Number of mole of Mg = 2.5/24 = 0.104mole
From the equation,
2moles of Mg required 1mole of O2.
Therefore, 0.104mole of Mg will require = 0.104/2 = 0.052mole of O2
Now let us calculate the volume occupy by 0.052mole of O2. This is shown below.
1mole of a gas occupy 22.4L at stp
Therefore, 0.052mole of O2 will occupy = 0.052 x 22.4 = 1.16L
Therefore, 1.16L of O2 is required for the reaction
