Answer: 61 grams
Explanation:
To calculate the number of moles, we use the equation:
The chemical equation for the combustion of octane in oxygen follows the equation:
By stoichiometry of the reaction;
25 moles of oxygen react with 2 moles of octane
4.69 moles of oxygen react with=
moles of octane
Thus, oxygen is the limiting reagent as it limits the formation of product and octane is the excess reagent.
25 moles of oxygen produce 18 moles of water
4.69 moles of oxygen produce=
moles of water.
Mass of water produced=
The maximum mass of water that could be produced by the chemical reaction is 61 grams.
Can’t help you man, sorry
Following reaction arise between Br2 and Cl2
Br2 + Cl2 → 2BrCl
(1mole) (1mole) (2moles)
From above balanced reaction, it can be seen that 1 mole of Br2 reacts with 1 mole of Cl2 to form 2 mole of BrCl
Thus, when <span>2.74 mol Cl2 reacts with excess Br2, 2.74 X 2 = 5.48 moles of BrCl will be formed. </span>
<span>To calculate the number of moles of aluminum, sulfur, and oxygen atoms in 4.00 moles of aluminum sulfate, al2(so4)3. We will simply inspect the "number" of aluminum, sulfur, and oxygen atoms available per one mole of the compound. Here we have Al2(SO4)3, which means that for every mole of aluminum sulfate, there are 2 moles of aluminum, 3 (1 times 3) moles of sulfur, and 12 (4x3) moles of oxygen. Since we have four moles of Al2(SO4)3 given, we simply multiply 4 times the moles present per 1 mole of the compound. So we have 4x2 = 8 moles of Al, 4x3 = 12 moles of sulfur, and 4x12 = 48 moles of oxygen.
So the answer is:
8,12,48
</span>
Answer:
The new concentration is 0.125 M.
Explanation:
Given data:
Initial volume V₁ = 125.0 mL
Initial molarity M₁ = 0.150 M
New volume V₂ = 25 mL +125 mL = 150 mL
New concentration M₂ = ?
Solution:
M₁V₁ = M₂V₂
0.150 M × 125 mL = M₂ × 150 mL
M₂ = 0.150 M × 125 mL / 150mL
M₂ = 18.75 M.mL/150 mL
M₂ = 0.125 M
The new concentration is 0.125 M.
