Stoichiometry problem on volume and concentration and stuff. I have no idea what to do

Question

Calculate the number of grams of SrCO3 made when 150 mL of 0.25 mol/L Na2CO3 reacts with excess SrCl2. The given equation is: SrCl2 + Na2CO3 = SrCO3 + 2NaCl. 

 

Please explain how to do it

Answer 1

 

In any stoichiometry problem we are comparing the amount of one substance in the reaction to the amount of another substance in the reaction using their molar ratios from the balanced chemical equation. If we are looking for the amount of product that can be formed (like we are in this problem because SrCO3 is a product), then we must compare it to the amount of limiting reactant because that's the reactant that will be used up first, thus limiting the amount of product that can be made. In this problem, it tells us that we have excess SrCl2, which means we are not worried about running out of it, so the other reactant, Na2CO3 must be limiting. So, overall, we are using the amount of Na2CO3 to determine how much SrCO3 can be formed. 

 

To determine the number of moles of  Na2CO3 present, we use the molarity equation and substitute in the information given, then rearrange it to solve for moles:

Molarity = moles/liters

0.25 M = moles/0.150 L

moles = (0.25 M) x (0.150 L) = 0.0375 moles Na2CO3

 

From there, we find the number of moles of SrCO3 by multiplying this value by the molar ratio of Na2CO3 to SrCO3 that we get from their coefficients in the balanced equation. They are in a 1:1 ratio here:

0.0375 moles Na2CO3 x (1 moles SrCO3/1 moles Na2CO3) = 0.0375 moles SrCO3

 

Finally, we calculate the mass of SrCO3 by multiplying the moles SrCO3 by its molar mass:

0.0375 moles SrCO3 x (147.63 g/mol) = 5.5 grams SrCO3