What is the limiting reactant of Sr(NO3)2+Na2CO3-->2NaNO3+SrCO3? Then what is the amount of strontium carbonate that can be made?(theoretical yield)
the two reactants both consume the same number of moles, so that will not determine a limiting reagent. On the other hand, if you have equal masses, then the one with the heavier molecular weight will have fewer moles, and thus limit the reaction.
To determine the limiting reactant in a chemical reaction, you need to compare the moles of each reactant to the stoichiometric ratio in the balanced equation.
Let's start by writing the balanced equation for the reaction:
Sr(NO3)2 + Na2CO3 → 2NaNO3 + SrCO3
The stoichiometric ratio tells us that 1 mole of Sr(NO3)2 will react with 1 mole of Na2CO3 to produce 1 mole of SrCO3.
Now, let's calculate the number of moles for each reactant:
First, we need to know the molar masses of each compound:
- Sr(NO3)2: strontium nitrate: 1 (strontium) + 2 (nitrate) = 1 + 2 × (14.01 + 3 × 16.00) = 211.6 g/mol
- Na2CO3: sodium carbonate: 2 (sodium) + 1 (carbon) + 3 (oxygen) = 2 × 22.99 + 12.01 + 3 × 16.00 = 105.99 g/mol
Assuming you have given the masses of each reactant, you can convert the masses to moles using the molar masses:
Moles of Sr(NO3)2 = Mass of Sr(NO3)2 / Molar mass of Sr(NO3)2
Moles of Na2CO3 = Mass of Na2CO3 / Molar mass of Na2CO3
After finding the moles of each reactant, we can compare them to determine the limiting reactant.
The limiting reactant is the one that is completely consumed, thus limiting the amount of product formed. It is the reactant that produces the smaller amount of product based on the stoichiometric ratio.
To calculate the theoretical yield of SrCO3, you can use the moles of the limiting reactant and the stoichiometry of the balanced equation.
Theoretical yield of SrCO3 = Moles of limiting reactant × Molar mass of SrCO3
By determining the limiting reactant and calculating the theoretical yield of SrCO3, we can find the amount of strontium carbonate that can be made.
To identify the limiting reactant in a chemical reaction, you need to compare the amounts of reactants present and the stoichiometry of the reaction.
Let's determine the limiting reactant using the given equation: Sr(NO3)2 + Na2CO3 → 2NaNO3 + SrCO3
First, you need to know the amount (in moles) of each reactant you have. Let's assume:
- Sr(NO3)2: x moles
- Na2CO3: y moles
Now, let's consider the stoichiometry of the reaction. From the balanced equation, you can see that the mole ratio between Sr(NO3)2 and SrCO3 is 1:1. This means that for every 1 mole of Sr(NO3)2, 1 mole of SrCO3 is produced.
Next, determine the mole ratio between Na2CO3 and SrCO3. From the balanced equation, you can see that the mole ratio between Na2CO3 and SrCO3 is 1:1 as well.
Now, let's compare the amounts of reactants to see which one is the limiting reactant.
- For Sr(NO3)2, the mole ratio with SrCO3 is 1:1.
- For Na2CO3, the mole ratio with SrCO3 is 1:1.
Thus, the reactant with the lowest mole ratio compared to SrCO3 will be the limiting reactant.
To determine the theoretical yield (amount of strontium carbonate produced), you need to calculate the moles of the limiting reactant and convert it to mass using the molar mass of SrCO3.
Let's assume that x moles of Sr(NO3)2 and y moles of Na2CO3 are reacted.
Since the mole ratio between Sr(NO3)2 and SrCO3 is 1:1, the number of moles of SrCO3 produced will also be x.
Next, you need to convert the moles of SrCO3 to grams. Multiply the moles of SrCO3 by the molar mass of SrCO3 (which you can find in the periodic table).
The resulting value will be the theoretical yield of strontium carbonate in grams.