for each of the following equations,calculate the mass of each product formed if 25.0g of the reaction listed first reacts completely with the second.a. 2AgNO3+CaS04=Ag2S04+Ca(No3)2. b. 2AL+GHNO3=2AL(NO3)3+3H2.
Here is a worked example that shows you how to do all of these.
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To calculate the mass of each product formed in the given equations, we need to use the concept of stoichiometry. Stoichiometry is a mathematical relationship between the number of atoms or molecules involved in a chemical reaction.
Let's solve the equations one by one:
a) 2AgNO3 + CaSO4 → Ag2SO4 + Ca(NO3)2
In this equation, the reactants are 2 moles of AgNO3 and 1 mole of CaSO4. The products formed are 1 mole of Ag2SO4 and 1 mole of Ca(NO3)2.
To calculate the mass of each product, we need to know the molar masses of the compounds:
- AgNO3: 169.87 g/mol
- CaSO4: 136.14 g/mol
- Ag2SO4: 311.81 g/mol
- Ca(NO3)2: 164.09 g/mol
First, we calculate the moles of each reactant:
- Moles of AgNO3: 25.0 g / (169.87 g/mol) = 0.1472 mol
- Moles of CaSO4: 25.0 g / (136.14 g/mol) = 0.1836 mol
According to the balanced equation, the stoichiometry tells us that the ratio of AgNO3 to Ag2SO4 is 2:1.
So, if we have 0.1472 mol of AgNO3, we would expect to produce 0.1472 / 2 = 0.0736 mol of Ag2SO4.
Now, we can calculate the mass of each product:
- Mass of Ag2SO4: 0.0736 mol x (311.81 g/mol) = 22.94 g
Similarly, the stoichiometry of CaSO4 to Ca(NO3)2 is also 1:1.
So, if we have 0.1836 mol of CaSO4, we would expect to produce 0.1836 mol of Ca(NO3)2.
- Mass of Ca(NO3)2: 0.1836 mol x (164.09 g/mol) = 30.11 g
Therefore, the mass of each product formed in this reaction is:
- Ag2SO4: 22.94 g
- Ca(NO3)2: 30.11 g
b) 2Al + 3HNO3 → 2Al(NO3)3 + 3H2
In this equation, the reactants are 2 moles of Al and 6 moles of HNO3. The products formed are 2 moles of Al(NO3)3 and 3 moles of H2.
Using the molar masses:
- Al: 26.98 g/mol
- HNO3: 63.01 g/mol
- Al(NO3)3: 213.01 g/mol
- H2: 2.02 g/mol
Calculating the moles of each reactant:
- Moles of Al: 25.0 g / (26.98 g/mol) = 0.9255 mol
- Moles of HNO3: 25.0 g / (63.01 g/mol) = 0.3961 mol
According to the balanced equation, the stoichiometry tells us that the ratio of Al to Al(NO3)3 is 2:2 (or 1:1).
So, if we have 0.9255 mol of Al, we would expect to produce 0.9255 mol of Al(NO3)3.
Now, let's calculate the mass of each product:
- Mass of Al(NO3)3: 0.9255 mol x (213.01 g/mol) = 197.28 g
Similarly, the stoichiometry of HNO3 to H2 is 3:3 (or 1:1).
So, if we have 0.3961 mol of HNO3, we would expect to produce 0.3961 mol of H2.
- Mass of H2: 0.3961 mol x (2.02 g/mol) = 0.80 g
Therefore, the mass of each product formed in this reaction is:
- Al(NO3)3: 197.28 g
- H2: 0.80 g
By following these calculations based on the stoichiometry of the balanced equations and using molar masses, we can determine the mass of each product formed when 25.0 g of the given reactants react completely.