Use the balanced equation below to answer the following questions.
2Al(NO3)3 + 3Na2CO3 Al2(CO3)3(s) + 6NaNO3
- What is the ratio of moles of Al(NO3)3 to moles Na2CO3? (1 point)
- If 852.04 g Al(NO3)3 reacted with the 741.93 g Na2CO3, which would be the limiting reagent? Show your work.
- Use the limiting reagent to determine how many grams of Al2(CO3)3 should precipitate out in the reaction.
- If only 341.63 g of Al2(CO3)3 precipitate were actually collected from the reaction, what would the percent yield be?
can you just answer the question
a. 2 to 3
please how do you get the amount of precipitate
To answer these questions, we will use stoichiometry and the concept of limiting reactants.
1. Ratio of moles of Al(NO3)3 to moles Na2CO3:
The balanced equation states that 2 moles of Al(NO3)3 react with 3 moles of Na2CO3. Therefore, the ratio of moles of Al(NO3)3 to moles of Na2CO3 is 2:3.
2. Determine the limiting reagent:
To determine the limiting reagent, we need to compare the amount of each reactant given and convert it to moles.
a) Al(NO3)3:
Given mass of Al(NO3)3: 852.04 g
Molar mass of Al(NO3)3 = 213.0 g/mol (24 + (14 + 3*16) * 3)
Moles of Al(NO3)3 = 852.04 g / 213.0 g/mol = 4.0 mol
b) Na2CO3:
Given mass of Na2CO3: 741.93 g
Molar mass of Na2CO3 = 105.99 g/mol (2*23 + 12 + 3*16)
Moles of Na2CO3 = 741.93 g / 105.99 g/mol = 7.0 mol
Now, using the balanced equation, we can see that the ratio of moles of Al(NO3)3 to Na2CO3 is 2:3.
Comparing the moles of both reactants, we see that Al(NO3)3 has 4.0 mol and Na2CO3 has 7.0 mol. Since the stoichiometric ratio requires 2 moles of Al(NO3)3 for every 3 moles of Na2CO3, we can calculate the actual ratio:
Moles of Al(NO3)3 / Moles of Na2CO3 = 4.0 mol / 7.0 mol = 0.57 (rounded to two decimal places)
Since the ratio is less than 2:3, Al(NO3)3 is the limiting reagent in this reaction.
3. Grams of Al2(CO3)3 precipitated:
To find the mass of Al2(CO3)3 formed, we need to use the molar ratio between Al(NO3)3 and Al2(CO3)3 from the balanced equation.
Molar mass of Al2(CO3)3 = 233.99 g/mol (2*27 + (12 + 3*16) * 3)
Molar ratio between Al(NO3)3 and Al2(CO3)3 = 2:1
Since Al(NO3)3 is the limiting reagent, we can calculate the moles of Al2(CO3)3 formed.
Moles of Al2(CO3)3 = Moles of Al(NO3)3 / Molar ratio = 4.0 mol / 2 = 2.0 mol
Mass of Al2(CO3)3 = Moles of Al2(CO3)3 * Molar mass = 2.0 mol * 233.99 g/mol = 468.0 g
Therefore, 468.0 grams of Al2(CO3)3 should precipitate out in the reaction.
4. Percent yield of Al2(CO3)3:
Percent yield = (Actual yield / Theoretical yield) * 100
Theoretical yield = 468.0 g (calculated in the previous step)
Actual yield = 341.63 g (given in the question)
Percent yield = (341.63 g / 468.0 g) * 100 = 73.1% (rounded to one decimal place)
Therefore, the percent yield of Al2(CO3)3 is 73.1%.