Determine the type of chemical reaction, find the limiting reagent, determine the amount of product in grams, and the amount in grams of excess reagent.
8.75 g of mercury (II) nitrate solution is mixed with 9.83 g sodium iodide solution (ppt).
Hg(NO3)2(aq) + 2NaI(aq) --> HgI2(s) + 2NaNO3(aq)
Chemical reaction: double replacement
Molar mass Hg(NO3)2 = 325amu. 1moleHg(NO3)2 = 325gHg(NO3)2
(1moleHg(NO3)2 / 325gHg(NO3)2) 8.75gHg(NO3)2 = .0269molesHg(NO3)2
Molar mass NaI = 150amu. 1moleNaI = 150gNaI
(1moleNaI / 150gNaI) 9.83gNaI = .0655molesNaI
For every 1 mole Hg(NO3)2, there are 2 moles Hg(NO3)2
Limiting Reagent: Hg(NO3)2
Product = 475g
I am not sure if I am figuring out the number of moles or grams in this problem. Also, I am not quite sure how to determine the limiting reagent. Please check/correct my work so I can get a better understanding on how to do this. Thank you!
you are ok (I didn't check math) until here:
For every 1 mole Hg(NO3)2, there are 2 moles Hg(NO3)2 ****That is nuts
Limiting Reagent: Hg(NO3)2***correct
Product = 475g****Nuts. You get .0269 moles of HgI2, that is the product. figure the mass of that.
Your approach to determining the number of moles and grams is correct. You have correctly calculated the number of moles of Hg(NO3)2 and NaI based on their respective molar masses.
To determine the limiting reagent, you need to compare the moles of each reagent to the stoichiometry of the balanced chemical equation. The balanced equation tells us that the ratio of Hg(NO3)2 to HgI2 is 1:1, and the ratio of NaI to HgI2 is 2:1.
Based on the stoichiometry, for every mole of Hg(NO3)2, we need 1 mole of HgI2. From the given moles, we have 0.0269 moles of Hg(NO3)2, which means we would have 0.0269 moles of HgI2 formed.
Similarly, for every mole of NaI, we need 2 moles of HgI2. From the given moles, we have 0.0655 moles of NaI, which means we would have 0.0328 moles of HgI2 formed.
Comparing the quantities of HgI2 formed from each reactant, we can see that we would have a smaller amount of HgI2 formed from Hg(NO3)2, which indicates that Hg(NO3)2 is the limiting reagent.
To determine the amount of HgI2 formed in grams, you can use the molar mass of HgI2, which is 454 g/mol. Since 1 mole of HgI2 is formed from 1 mole of Hg(NO3)2, we have 0.0269 moles of HgI2. Multiplying the mole amount by the molar mass, we find that the amount of HgI2 formed is 12.24 grams.
To find the amount of excess reagent, we can subtract the moles of the limiting reagent consumed from the moles of the excess reagent. In this case, the limiting reagent is Hg(NO3)2. Therefore, to find the moles of excess NaI, we subtract the moles of HgI2 formed from the moles of NaI.
0.0655 moles NaI - 0.0328 moles HgI2 = 0.0327 moles NaI (excess)
To find the amount of excess NaI in grams, we can multiply the moles by the molar mass of NaI, which is 150 g/mol.
0.0327 moles NaI * 150 g/mol = 4.90 grams NaI (excess)
To determine the limiting reagent and the amount of product in grams, you need to calculate the number of moles of each reactant and compare them to the stoichiometry of the balanced chemical equation. Here's how you can do it step by step:
1. Calculate the number of moles for each reactant:
- For Hg(NO3)2:
- Molar mass Hg(NO3)2 = 200.59 g/mol (note: you provided an incorrect molar mass)
- Moles of Hg(NO3)2 = 8.75 g / 200.59 g/mol = 0.0436 mol
- For NaI:
- Molar mass NaI = 149.89 g/mol
- Moles of NaI = 9.83 g / 149.89 g/mol = 0.0656 mol
2. Determine the limiting reagent:
- The balanced equation shows that it requires a 1:2 mole ratio of Hg(NO3)2 to NaI. Therefore, the limiting reagent will be the one with a smaller number of moles per mole ratio.
- In this case, since 0.0436 mol of Hg(NO3)2 is less than 0.0656 mol of NaI, Hg(NO3)2 is the limiting reagent.
3. Calculate the amount of product in grams:
- From the balanced equation, the mole ratio between Hg(NO3)2 and HgI2 is 1:1. This means that for every 1 mole of Hg(NO3)2, you will get 1 mole of HgI2.
- Therefore, the moles of HgI2 produced will be the same as the moles of Hg(NO3)2 used, which is 0.0436 mol.
- Molar mass HgI2 = 454.38 g/mol
- Mass of HgI2 = 0.0436 mol × 454.38 g/mol = 19.8 g
4. Calculate the amount in grams of the excess reagent:
- To determine the excess reagent, you can subtract the moles of the limiting reagent used from the total moles of the corresponding reactant.
- Since Hg(NO3)2 is the limiting reagent, the moles of NaI used can be calculated as follows:
- Moles of NaI used = 0.0436 mol × (2 mol NaI / 1 mol Hg(NO3)2) = 0.0872 mol
- The excess moles of NaI can be calculated as follows:
- Excess moles of NaI = 0.0656 mol - 0.0872 mol = -0.0216 mol
(a negative sign indicates an excess of the reactant)
- To convert the excess moles of NaI to grams, use the molar mass of NaI:
- Mass of excess NaI = -0.0216 mol × 149.89 g/mol = -3.23 g
(the negative sign indicates an excess)
So, to summarize:
- The type of chemical reaction is a double replacement reaction.
- The limiting reagent is Hg(NO3)2.
- The amount of HgI2 produced is 19.8 grams.
- The amount of excess NaI is -3.23 grams. Note: A negative sign indicates an excess amount.
Please note that there was a correction in the molar mass of Hg(NO3)2, which should be 200.59 g/mol instead of 325 g/mol.