The percentage yield of the product as obtained is 73%
The limiting reactant is the reactant that is present in the least amount in the reaction.
Given the reaction;2Al(NO3)3 + 3Na2CO3 → Al2(CO3)3 + 6NaNO3, the mole ratio of Al(NO3)3 to Na2CO3 is 2:3.
Number of moles of Na2CO3 = 741.93 g/106 g/mol = 7 moles
Number of moles of Al(NO3)3 = 852.04 g /213 g/mol = 4 moles
If 2 moles of Al(NO3)3 reacts with 3 moles of Na2CO3
4 moles of Al(NO3)3 reacts with 4 moles * 3 moles /2 moles
= 6 moles
Hence Al(NO3)3 is the limiting reactant
2 moles of Al(NO3)3 produces 1 moles of Al2(CO3)3
4 moles of Al(NO3)3 produces 4 moles * 1 moles/ 2 moles
= 2 moles of Al2(CO3)3
Theoretical yield of Al2(CO3)3 = 2 moles * 234 g/mol = 468 g
Percent yield = 341.63 g/468 g * 100/1
= 73%
Use the balanced equation below to answer the following questions. 2Al(NO3)3 + 3Na2CO3 >>>>> Al2(CO3)3(s) + 6NaNO3 a. What is the ratio of moles of Al(NO3)3 to moles Na2CO3? b. If 852.04 g Al(NO3)3 reacted with the 741.93 g Na2CO3, which would be the limiting reagent? Show your work.c. Use the limiting reagent to determine how many grams of Al2(CO3)3 should precipitate out in the reaction. d. If only 341.63 g of Al2(CO3)3 precipitate were actually collected from the reaction, what would the percent yield be?
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