Here is the equation:
Ca(OH)2 + CO2 --------> CaCO3 + H2O
2.How many grams of CaCO3 would be formed if 4.75g of Ca(OH)2 were allowed to react with 3.25g of CO2?
10 points will be rewarded!!
Ca(OH)2 + CO2 --------> CaCO3 + H2O
2.How many grams of CaCO3 would be formed if 4.75g of Ca(OH)2 were allowed to react with 3.25g of CO2?
10 points will be rewarded!!
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You are correct in stating you need to find the limiting reagent or the reagent in excess.
You need to work out the number of moles you have of both reactants
n=m/M n = number of moles, m = mass (g) , M = molar mass (g mol-1)
so for Ca(OH)2
n= (4.75)/(40+16+16+1+1) = 0.0642 moles
For CO2
n (3.25)/(12+16+16)= 0.0739 moles
So its clear to see that you have CO2 in excess and that Ca(OH)2 is your limiting reagent.
looking at the stoichiometry of your equation Ca(OH)2 + CO2 --------> CaCO3 + H2O
its clear to see that one mole of Ca(OH)2 will form one mole of CaCO3 (1:1 ratio)
you do not have one mole, you have 0.0642 moles of Ca(OH)2 so the maximum amount of CaCO3 that could be formed will also be 0.0642 moles.
using the equation n=m/M we can re-arrange it to make m the subject:
m=nM so to work out the mass of CaCO3
m = (0.0642)(40+12+16+16+16) = 6.42g of CaCO3 is the maximum mass you could produce.
That fully answers your question but here is a little thing i like to do just to help you understand it more. In total your reactants weighed 8.00g (3.25g of CO2 and 4.75g of Ca(OH)2) and we worked out we would form 6.42 g of CaCO3. so where is the other 1.58 g?
well it is in the excess CO2 left over and the H2O formed. If we look back we will recall that the stoichiometry is 1:1 ratio throughout the reaction. We worked out the number of moles of CO2 we had (0.0739 moles) if this reacts in a 1:1 ratio with Ca(OH)2 then 0.0642 moles of CO2 is used up in the reaction. This leaves (0.0739-0.0642) 0.0097 moles of CO2 in excess, This in a mass is :
m= (0.0097)(12+16+16) = 0.4268 g
Now the water, we will produce 0.0642 moles of H2O in the reaction this in mass :
m = (0.0642)(16+1+1) = 1.1556g
SO overall we have :
6.42g of CaCO3 produced
1.1556g H2O produced
0.4268 g CO2 excess
6.42+1.1556+0.4268 = 8.00 g which is the mass of the reactants at the start.
This little method helps you double check you answers and also aids in understand how it all works.
You need to work out the number of moles you have of both reactants
n=m/M n = number of moles, m = mass (g) , M = molar mass (g mol-1)
so for Ca(OH)2
n= (4.75)/(40+16+16+1+1) = 0.0642 moles
For CO2
n (3.25)/(12+16+16)= 0.0739 moles
So its clear to see that you have CO2 in excess and that Ca(OH)2 is your limiting reagent.
looking at the stoichiometry of your equation Ca(OH)2 + CO2 --------> CaCO3 + H2O
its clear to see that one mole of Ca(OH)2 will form one mole of CaCO3 (1:1 ratio)
you do not have one mole, you have 0.0642 moles of Ca(OH)2 so the maximum amount of CaCO3 that could be formed will also be 0.0642 moles.
using the equation n=m/M we can re-arrange it to make m the subject:
m=nM so to work out the mass of CaCO3
m = (0.0642)(40+12+16+16+16) = 6.42g of CaCO3 is the maximum mass you could produce.
That fully answers your question but here is a little thing i like to do just to help you understand it more. In total your reactants weighed 8.00g (3.25g of CO2 and 4.75g of Ca(OH)2) and we worked out we would form 6.42 g of CaCO3. so where is the other 1.58 g?
well it is in the excess CO2 left over and the H2O formed. If we look back we will recall that the stoichiometry is 1:1 ratio throughout the reaction. We worked out the number of moles of CO2 we had (0.0739 moles) if this reacts in a 1:1 ratio with Ca(OH)2 then 0.0642 moles of CO2 is used up in the reaction. This leaves (0.0739-0.0642) 0.0097 moles of CO2 in excess, This in a mass is :
m= (0.0097)(12+16+16) = 0.4268 g
Now the water, we will produce 0.0642 moles of H2O in the reaction this in mass :
m = (0.0642)(16+1+1) = 1.1556g
SO overall we have :
6.42g of CaCO3 produced
1.1556g H2O produced
0.4268 g CO2 excess
6.42+1.1556+0.4268 = 8.00 g which is the mass of the reactants at the start.
This little method helps you double check you answers and also aids in understand how it all works.
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Ca(OH)2 has a molar mass of 74.09 g/mol, so 4.75g x (1 mol/74.09g) = 0.064 mol Ca(OH)2.
CO2 has a molar mass of 44.01 g/mol, so 3.25g x (1mol/44.01g) = 0.074 mol CO2.
CO2 would be in excess, aka Ca(OH)2 would be the limiting reactant b/c less moles of it react therefore the same amount of moles of CO2 would have to react. The one with the smaller number of moles is the limiting reactant and the larger # of moles is in excess
hope that helps!
CO2 has a molar mass of 44.01 g/mol, so 3.25g x (1mol/44.01g) = 0.074 mol CO2.
CO2 would be in excess, aka Ca(OH)2 would be the limiting reactant b/c less moles of it react therefore the same amount of moles of CO2 would have to react. The one with the smaller number of moles is the limiting reactant and the larger # of moles is in excess
hope that helps!