Answer: Option (c) is the correct answer.
Explanation:
A limiting reagent is defined as a reagent that completely gets consumed in a chemical reaction. A limiting reagent limits the formation of products.
For example, we have given 5 mol of A and the reaction is 
Whereas when 4 mol B will react with 2 mol of A. Hence, 8 mol of B will react with 4 mol A as follows.
= 4 mol
As, the given moles of A is more than the required moles. Thus, it is considered as an excess reagent.
Hence, B is a limiting reagent because it limits the formation of products.
Thus, we can conclude that limiting reactant is the term used to describe the reactant that is used up completely and controls the amount of product that can be produced during a chemical reaction.
Answer:
1.09 L
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the volume in liters of a 0.360 mol/L barium acetate solution that contains 100 g of barium acetate. Be sure your answer has the correct number of significant digits.</em>
<em />
The molar mass of barium acetate is 255.43 g/mol. The moles corresponding to 100 grams are:
100 g × (1 mol/255.43 g) = 0.391 mol
0.391 moles of barium acetate are contained in an unknown volume of a 0.360 mol/L barium acetate solution. The volume is:
0.391 mol × (1 L/0.360 mol) = 1.09 L
Answer:
39.7 %
Explanation:
magnesium + oxygen ⟶ magnesium oxide
10.57 g 6.96 g 17.53 g
According to the <em>Law of Conservation of Mass</em>, the mass of the product must equal the total mass of the reactants.
Mass of MgO = 10.57 + 6.96
Mass of MgO = 17.53 g
The formula for mass percent is
% by mass = Mass of component/Total mass × 100 %
In this case,
% O = mass of O/mass of MgO × 100 %
Mass of O = 6.96 g
Mass of MgO = 17.53 g
% O = 6.96/17.53 × 100
% O = 0.3970 × 100
% O = 39.7 %