Using balanced equations to work out the mass of a product from the mass of a reactant, and identifying which reactant limits the amount produced. Higher tier content.
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1.State the three steps for calculating a reacting mass.
Convert the known mass into moles, use the balanced equation to find the moles of the substance you want, then convert those moles back into a mass.
2.Why must the equation be balanced first?
The balancing numbers give the mole ratio, which is what links one substance to another.
3.In 2Mg + O2 gives 2MgO, how many moles of MgO come from 2 moles of Mg?
2 moles, because the ratio of Mg to MgO is 1 to 1.
4.Calculate the mass of MgO from 48 g of Mg. Mg is 24, MgO is 40.
Moles of Mg = 48 divided by 24 = 2. Ratio is 1 to 1, so 2 moles of MgO. Mass = 2 x 40 = 80 g.
5.In CaCO3 gives CaO + CO2, how many moles of CO2 come from 1 mole of CaCO3?
One, because the ratio is 1 to 1.
6.Calculate the mass of CaO from 50 g of CaCO3. CaCO3 is 100, CaO is 56.
Moles of CaCO3 = 50 divided by 100 = 0.5. Ratio is 1 to 1, so 0.5 moles of CaO. Mass = 0.5 x 56 = 28 g.
7.In N2 + 3H2 gives 2NH3, how many moles of ammonia come from 1 mole of nitrogen?
Two, because the ratio of N2 to NH3 is 1 to 2.
8.Calculate the mass of ammonia from 28 g of nitrogen. N2 is 28, NH3 is 17.
Moles of N2 = 28 divided by 28 = 1. Ratio 1 to 2, so 2 moles of NH3. Mass = 2 x 17 = 34 g.
9.What is a limiting reactant?
The reactant that is completely used up first, which therefore limits how much product can be made.
10.What is meant by a reactant being in excess?
There is more of it than is needed, so some is left over when the reaction stops.
11.How do you identify the limiting reactant?
Convert each reactant mass into moles, divide each by its balancing number, and the smallest value identifies the limiting reactant.
12.Why does the amount of product depend on the limiting reactant?
Once it is used up the reaction stops, no matter how much of the other reactants remain.
13.Why is one reactant often deliberately added in excess?
To make sure the more expensive or more important reactant is completely used up rather than wasted.
14.In 2Mg + O2, you have 3 moles of Mg and 3 moles of O2. Which limits the reaction?
Magnesium. 3 moles of Mg needs only 1.5 moles of O2, so oxygen is in excess and magnesium runs out first.
15.What happens to the mass of product if you double the limiting reactant?
It doubles, provided the other reactants are still in excess.
16.What happens to the mass of product if you double a reactant already in excess?
Nothing. The limiting reactant still controls how much can be made.
17.What does the gradient of a mass against time graph show?
The rate at which the product is forming. A steeper line means a faster reaction.
18.What does it mean when a mass against time graph levels off?
The reaction has finished because the limiting reactant has been completely used up.
19.Two experiments level off at the same mass but at different times. What does that tell you?
They had the same amount of limiting reactant, but one reaction was faster than the other.
20.Calculate the mass of hydrogen from 24 g of magnesium in Mg + 2HCl gives MgCl2 + H2. Mg is 24, H2 is 2.
Moles of Mg = 24 divided by 24 = 1. Ratio 1 to 1, so 1 mole of H2. Mass = 1 x 2 = 2 g.
21.Why might the mass you actually obtain be less than your calculation predicts?
Some product is lost during transfer or purification, the reaction may not go to completion, or side reactions may occur.
22.What must you always check before using an equation for a mass calculation?
That it is correctly balanced, because the mole ratio depends entirely on that.
23.How many moles of oxygen are needed to burn 1 mole of methane in CH4 + 2O2 gives CO2 + 2H2O?
Two moles.
24.If a calculation gives 0.35 moles of a substance with Mr 40, what is the mass?
0.35 x 40 = 14 g.
25.Why does converting to moles make comparisons possible?
Moles count particles. Masses cannot be compared directly because different substances have different particle masses.
26.A reaction produces 1.5 moles of a gas with Mr 44. Calculate the mass.
1.5 x 44 = 66 g.
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