If 100. mL of 0.400 M Na2SO4 is added to 200. mL of 0.600 M NaCl, what is the concentration of Na+ ions in the final solution? A
1 answer:
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Answer:
Y = 92.5 %
Explanation:
Hello there!
In this case, since the reaction between lead (II) nitrate and potassium bromide is:
Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:
Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:
And the resulting percent yield:
Regards!