Chemistry Test - Quantities in Chemical Reactions and Solutions Flashcards
- What is the best description of a Limiting Reagent?
- The reagent that is the first one in the reaction equation
- The reagent that is left over when the reaction is done
- The reagent that has the smallest Molar coefficient in front
- The reagent that is a solid, since solids always run out
- The reagent that will run out first stopping the reaction
Answer:
- The reagent that will run out first, stopping the reaction.
The best description of the limiting reagent is option E. The limiting reagent is the reactant that is consumed completely or used up first in a chemical reaction. Once the limiting reagent is completely consumed, the reaction cannot proceed any further because there are no more molecules available for the reaction to continue. The limiting reagent determines the maximum amount of product that can be formed in the reaction.
- In the following reaction, If 2 moles of Ca(OH)) Is needed to react with all of the HyPO, that a student has, but only 1.5 moles of Ca(OH). are available, what is true about the Ca(OHh.?
- The student needs more Ca(OH).., than they have meaning it is Limiting
- The student has more Ca(OH). than needed meaning it is in excess
- The student needs more HPO and should order more
- The Ca(OH), is present in stoichiometric amounts (just the right amount)
- The student has less Ca(OH) than needed meaning it is in excess.
Answer:
- The student needs more Ca(OH)2 than they have, meaning it is limiting.
In the given scenario, the student has 1.5 moles of Ca(OH)2 available, but 2 moles of Ca(OH)2 are needed to react with all of the HPO. Since the student has less Ca(OH)2 than is required for the complete reaction, Ca(OH)2 is the limiting reagent. The limiting reagent determines the maximum amount of product that can be formed, and in this case, the reaction will be limited by the available amount of Ca(OH)2. Therefore, option A is correct
- What is the mole ratio for Ca(OH)/H,PO.?
- 2/1
- ½
- 3/2
- ⅔
- 1.5/3
Answer:
- 3/2
- What is the stoichiometic amount (just the right amount) of HO, needed to react with 1.5 moles of Ca(OH)a from question #2?
- 2.28 moles of H,PO, are needed to react with 1.5 moles Ca(OH)2ia)
- 4.8 moles of HyPO, are needed to react with 1.5 moles Ca(OH)28)
- 3.0 moles of H,PO, are needed to react with 1.5 moles Ca(OH)24)
- 1.0 moles of HyPO, are needed to react with 1.5 moles Ca(OH)2»
- 1.5 moles of HyPO, are needed to react with 1.5 moles Ca(OH)20)
Answer:
- 1.0 moles of H3PO4 are needed to react with 1.5 moles of Ca(OH)2.
- When diluting a dark purple solution by adding water, what will happen to the colour?
- The colour will become a darker purple
- The colour is not change
- The colour will become more blue
- The colour will become a lighter colour
- The colour will disappear completely even if only a little water is added
Answer:
- The color will become a lighter color.
When diluting a dark purple solution by adding water, the most likely outcome is that the color will become lighter.
As water is added to the solution, it will decrease the concentration of the solute (the purple substance) in the solution. This decrease in concentration leads to a decrease in the intensity of the color, resulting in a lighter shade of purple. However, it is important to note that the exact change in color may depend on the specific solute and its concentration in the solution.
6 When dealing with dilutions n = CV and the dilution equation is C1V1 = C2V2 - Why is that the dilution equation?
- The Volume of water always decreases as the concentration is lowered.
- The moles are the same since only water was added, and no chemicals are removed
- The concentration always increases as more water is added
- The concentration of the solution before and after the dilution is the same
- The dilution equation actually doesn’t work and I should ask this question
Answer:
- The moles are the same since only water was added, and no chemicals are removed.
- Why does the concentration go down when water is added?
- The addition of water reduces the amount of chemicals present
- The addition of water increases the amount of chemicals present
- The same amount of chemicals are spread out over less water
- The same amount of chemicals are spread out over more water
- All of the above
Answer:
- The same amount of chemicals are spread out over more water.
When water is added to a solution, the volume of the solution increases while the amount of solute (chemicals) remains the same. As a result, the solute becomes more spread out or diluted within the larger volume of water.
- What is the solute in a solution?
- The substance is dissolved (Le., a chemical)
- The substance that reacts with water.
- A substance that does not dissolve well (solute is another word for precipitate)
- The substance that does the dissolving (Le., water)
- None of the above
Answer:
- The substance is dissolved (i.e., a chemical)
In a solution, the solute refers to the substance that is dissolved in the solvent to form a homogeneous mixture. The solute can be a solid, liquid, or gas that becomes dispersed at the molecular or ionic level within the solvent.
- What are the units that we use for concentration when using the formula C = n/V?
- L/mol
- mol/l
- ml/mol
- g/L
- mol/ml
Answer:
- mol/L
When using the formula C = n/V, where C represents concentration, n represents the amount of substance (in moles), and V represents volume (in liters), the units of concentration are typically expressed as moles per liter (mol/L). This unit represents the amount of solute (in moles) present in one liter of the solution.
- What is the concentration of a solution if 7.8 moles of Ba(OH). are dissolved in 3300 ml of water in mol/L?
- 2.36
- 434
- 0.0025
- 45.70
- 7.8
Answer:
- 2.36
C = n/V
3300 ml = 3300/1000 = 3.3 L
C = 7.8 moles / 3.3 L ≈ 2.36 mol/L
Therefore, the concentration of the solution is approximately 2.36 mol/L.
- For the following reaction, determine the MASS of CO2(g) produced if 18g of Oxygen is used to burn in excess of Butane (C4H10).
Step 1: Write down all the known masses with - in excess, 18g, x, and n/a
Step 2: Calculate all the molar masses
Step 3: Use the mass and molar mass of oxygen to calculate the number of moles (n=m/MM)
Step 4: Use the mole ratio of O2 and CO2 (13/8) and (0.5/x) to find the moles in CO2
Step 5: Use the number of moles and molar mass to find the mass of CO2 [(n)(MM) = M]
- Calculate the percentage yield if only 42g of Ag is produced.
Step 1: Write down the givens as the theoretical yield and actual yield
Step 2: Write down the required as “% yield = ?”
Step 3: The analysis is found using the formula “Actual yield/theoretical yield x 100”
Step 4: Solve and write the phrase
- In the following reaction, 0.5 mol of Al are reacted with 1.6 mol of HCl, which one is the limiting reagent?
Step 1: Write down the given moles in the molecular formula
Step 2: Write down the required as “Limiting reagent ?”
Step 3: Use the mole ratio of Al and HCl to figure out how many moles there SHOULD BE for HCl by replacing the moles in the ratio with x as a variable
Step 4: Solve and learn that there’s more HCl than needed, making it the one in excess and Al the one that’s limiting.
- What is the molar concentration of a KOH solution (MM = 56.11 g/mol) in which 5g of solid KOH is dissolved in 80 ml of water?
Step1: Write down givens and make sure to convert the volume from ml to L by doing (ml/1000L)
Step 2: Write down required as NKOH = ? and CKOh = ?
Step 3: Use the mass and molar mass to find the mols (n) in KOH (n=m//MM)
Step 4: Use the formula c = n/v to find concentration
- A student wants to dilute a 0.85 mol/L solution of Potassium Permanganate to make 57 ml of a 0.2 mol/L solution. What volume of the 0.85 mol/L solution does the student need to measure out?
Step 1: Write down the givens as everything but v2 and make sure to convert the volume from ml to L by doing (ml/1000L)
Step 2: Write down the formula c1v1 = c2v2
Step 3: Fill in everything but V2 and rearrange and solve for the unknown
