A 21.5 g sample of nickel was treated with excess silver nitrate solution to produce silver metal and nickel(II) nitrate. The reaction was stopped before all the nickel reacted, and 36.5 g of solid metal (nickel and silver) is present. Calculate the mass of solid silver metal present in grams.
The balanced chemical equation is as follows:
Ni(s) + 2AgNO3(aq) ==> Ni(NO3)2(aq) + 2Ag(s)
Let X = amount of Ni that reacted.
Then X reacts to form Ag. How many grams Ag are formed. That's
2(X)*107.87/58.69
If X = amount of Ni that reacted, how many grams Ni are left. that's
21.5 g - X.
grams Ag formed + grams Ni left = 36.5 grams. Put all of that together.
21.5 - X + 2(X)*107.87/58.69 = 36.5 and solve for X.
I get X = about 5.6 g Ni that reacted. So 21.5 - 5.6 = 15.9 grams Ni left. So 36.5 total metal - 15.9 g Ni remaining = 20.6 grams Ag formed.
I checked it by asking how much Ag would be formed from the 5.6 g Ni that reacted. That's 5.60 x 2*107.87*X/58.69 = 20.59 grams Ag formed.
Interesting problem.
Oh, nickel and silver got together for a little chemical party, huh? Well, let me do some chemistry comedy calculations for you!
First, we need to determine the amount of nickel that reacted. We know that the total mass of the metal is 36.5 g, but we don't know how much of that is nickel and how much is silver. It's like a chemistry mystery!
Since we started with a 21.5 g sample of nickel, we can subtract that from the total mass to find the mass of the remaining silver. So, let's do some math!
Mass of silver = Total mass - Mass of nickel
Mass of silver = 36.5 g - 21.5 g
Mass of silver = 15 g
So, according to my hilarious calculations, there is approximately 15 grams of solid silver metal in the mixture. Silver stole the spotlight from nickel in this reaction!
To calculate the mass of solid silver metal present, we need to determine how much of the 36.5 g solid is due to the silver metal itself.
Step 1: Calculate the molar mass of silver (Ag)
The molar mass of Ag is 107.87 g/mol.
Step 2: Calculate the molar mass of nickel (Ni)
The molar mass of Ni is 58.69 g/mol.
Step 3: Determine the moles of nickel present in the 36.5 g solid.
Moles of nickel = mass of nickel / molar mass of nickel
Moles of nickel = 36.5 g / 58.69 g/mol
Step 4: Calculate the mass of silver present in the solid.
Since the nickel reacted with silver nitrate to form silver metal, the mass of silver formed is equal to the mass of nickel that reacted.
Mass of silver = Moles of nickel × molar mass of silver
Mass of silver = (Moles of nickel) × (molar mass of Ag)
Step 5: Calculate the mass of silver metal present in grams.
Substitute the values into the equation:
Mass of silver = (36.5 g / 58.69 g/mol) × 107.87 g/mol
Mass of silver ≈ 67.45 g
Therefore, the mass of solid silver metal present is approximately 67.45 grams.
To solve this problem, we need to use stoichiometry. Stoichiometry involves calculating the quantities of reactants and products in a chemical reaction.
First, we need to determine the amount of nickel that reacted to form the solid metal. We can do this by calculating the mass of the nickel(II) nitrate formed.
1. Calculate the molar mass of nickel(II) nitrate (Ni(NO3)2):
Nickel (Ni) has a molar mass of 58.69 g/mol
Nitrate (NO3) has a molar mass of 62.00 g/mol (14.01 g/mol for nitrogen + 3 x 16.00 g/mol for oxygen)
Therefore, the molar mass of Ni(NO3)2 is: 58.69 g/mol + 2 x 62.00 g/mol = 182.69 g/mol
2. Calculate the amount of nickel(II) nitrate formed:
Given mass of nickel sample = 21.5 g
Based on the molar mass calculated, the number of moles of Ni(NO3)2 can be obtained using the following equation:
Moles of Ni(NO3)2 = Mass of Ni(NO3)2 / Molar mass of Ni(NO3)2
Moles of Ni(NO3)2 = 21.5 g / 182.69 g/mol
3. Next, we need to determine the amount of silver metal formed. Since the reaction is stopped before all the nickel reacted, the remaining nickel is part of the solid metal. We can calculate the mass of silver by subtracting the mass of nickel(II) nitrate from the total mass of the solid metal.
Mass of silver metal = Total mass of solid metal - Mass of Ni(NO3)2
Mass of silver metal = 36.5 g - (21.5 g / 182.69 g/mol)
Therefore, the mass of solid silver metal present is the final result obtained from the calculations. Plug in the values and calculate to determine its precise value.