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.
Calcium nitrate is decomposed with heat to give calcium nitrite and oxygen gas. Start with 15.99g of calcium nitrate.
Ca(NO3)2(aq) -heat-> Ca(NO3)2(s) + O2(g)
chemical reaction is decomposition.
Molar Mass of Ca(NO3)2 = 164 amu. 1moleCa(NO3)2 = 164gCa(NO3)2
(1moleCa(NO3)2 / 164gCa(NO3)2) 15.99gCa(NO3)2 = .0975moleCa(NO3)2
There is no limiting reagent.
The product would be 15.99g.
I don't think this is right and I don't know how to answer the last question (especially in this problem).
Could you please explain this problem and check my work? Thanks!
Sure! Let's go through the problem step by step and check your work.
1. Determine the type of chemical reaction:
The given reaction is Ca(NO3)2(aq) -heat-> Ca(NO3)2(s) + O2(g). This reaction is a decomposition reaction because a single reactant, calcium nitrate (Ca(NO3)2), breaks down into multiple products, calcium nitrite (Ca(NO3)2) and oxygen gas (O2).
2. Find the limiting reagent:
To determine the limiting reagent, we need to compare the moles of each reactant present to the stoichiometric ratio in the balanced equation. In this case, the equation is balanced, which means 1 mole of Ca(NO3)2 produces 1 mole of Ca(NO3)2 and 1 mole of O2.
You correctly calculated the moles of Ca(NO3)2 to be 0.0975 moles based on the given mass of 15.99g. However, it seems you made a mistake in your statement that there is no limiting reagent. To determine if there is a limiting reagent, you need to compare the moles of the reactants in the given ratio.
Can you provide the molar mass of O2? With that information, we can calculate the moles of oxygen gas (O2) produced.
3. Determine the amount of product in grams:
To find the amount of Ca(NO3)2(s) produced, we first need to determine the mole ratio of Ca(NO3)2(s) to Ca(NO3)2(aq) using the balanced equation. The coefficient ratio is 1:1, meaning 1 mole of Ca(NO3)2(aq) will produce 1 mole of Ca(NO3)2(s).
You correctly converted the mass of Ca(NO3)2 to moles, which is 0.0975 moles. So, the amount of Ca(NO3)2(s) produced will also be 0.0975 moles.
To calculate the mass of Ca(NO3)2(s), we multiply the moles by the molar mass of Ca(NO3)2: (0.0975 moles) x (164 g/mol) = 15.99 grams.
So, you were correct in stating that the mass of the product (Ca(NO3)2(s)) is 15.99 grams.
4. Determine the amount of excess reagent in grams:
To find the excess reagent, we need to compare the moles of the reactants present. Can you provide the molar mass of O2 so that we can calculate the moles of oxygen gas (O2) produced? Once we have that information, we can proceed with calculating the amount of excess reagent.
Please let me know the molar mass of O2, and I'll be able to help you further.