For each of the following precipitation reactions calculate how many grams of the first reactant are necessary to completely react with 55.8 g of the second reactant. K2SO4 + Sr(NO3)2 + SrSO4 + 2KNO3
I think you mean
K2SO4(aq) + Sr(NO3)2(aq) > SrSO4 (s)+ 2KNO3 (aq)
so you need one mole of the first reactant for each mole of the second reactant.
How many moles of strontium nitrtate=55.8/211.6
so you need that number of moles of postassium sulfate, and in grams that is
55.8/211.6 * formulamassK2SO4=
55.8/211.6 *174.3 grams
Why did the chemist have a dog? Because he wanted a lab assistant! Now, let's solve this chemistry problem together.
To find out how many grams of the first reactant (K2SO4) are needed to completely react with 55.8 g of the second reactant (Sr(NO3)2), we'll need to determine the balanced equation for the reaction and use stoichiometry.
The balanced equation for this precipitation reaction is:
K2SO4 + Sr(NO3)2 -> SrSO4 + 2KNO3
From the balanced equation, we can see that the stoichiometric ratio of K2SO4 to Sr(NO3)2 is 1:1. This means that 1 mole of K2SO4 will react with 1 mole of Sr(NO3)2.
First, we need to find the molar mass of Sr(NO3)2:
Sr(NO3)2 = (1 x Sr) + (2 x N) + (6 x O)
Sr(NO3)2 = 1 x 87.6 + 2 x 14.01 + 6 x 16.00
Sr(NO3)2 = 211.58 g/mol
Now we can calculate the number of moles of Sr(NO3)2:
55.8 g / 211.58 g/mol = 0.264 mol of Sr(NO3)2
Since the stoichiometric ratio of K2SO4 to Sr(NO3)2 is 1:1, we know that we need 0.264 moles of K2SO4.
Finally, let's find the mass of K2SO4:
Mass of K2SO4 = (0.264 mol of K2SO4) x (174.3 g/mol of K2SO4)
Mass of K2SO4 = 45.94 g
So, approximately 45.94 grams of K2SO4 are necessary to completely react with 55.8 grams of Sr(NO3)2.
To calculate the grams of the first reactant necessary to react with 55.8 g of the second reactant, we need to follow these steps:
1. Write the balanced chemical equation for the precipitation reaction:
K2SO4 + Sr(NO3)2 -> SrSO4 + 2KNO3
2. Find the molar masses of the compounds involved:
- K2SO4: 2(39.10 g/mol K) + 32.06 g/mol S + 4(16.00 g/mol O) = 174.26 g/mol
- Sr(NO3)2: 1(87.62 g/mol Sr) + 2(14.01 g/mol N) + 6(16.00 g/mol O) = 211.63 g/mol
- SrSO4: 1(87.62 g/mol Sr) + 1(32.06 g/mol S) + 4(16.00 g/mol O) = 183.68 g/mol
- 2KNO3: 2(39.10 g/mol K) + 2(14.01 g/mol N) + 6(16.00 g/mol O) = 194.19 g/mol
3. Determine the stoichiometric ratio between the reactants:
From the balanced equation, we can see that 1 mole of K2SO4 reacts with 1 mole of Sr(NO3)2 to produce 1 mole of SrSO4 and 2 moles of KNO3.
4. Convert the given mass of the second reactant (Sr(NO3)2) to moles:
moles = mass / molar mass
moles of Sr(NO3)2 = 55.8 g / 211.63 g/mol = 0.2639 mol
5. Use the stoichiometric ratio to find the moles of the first reactant (K2SO4) required:
According to the stoichiometry, 1 mole of K2SO4 reacts with 1 mole of Sr(NO3)2.
Therefore, moles of K2SO4 = 0.2639 mol
6. Convert moles of K2SO4 to grams:
grams = moles × molar mass
grams of K2SO4 = 0.2639 mol × 174.26 g/mol ≈ 45.97 g (rounded to two decimal places)
Therefore, approximately 45.97 grams of K2SO4 are necessary to completely react with 55.8 grams of Sr(NO3)2 in the given precipitation reaction.
To determine the grams of the first reactant required to completely react with the given amount of the second reactant, you need to balance the equation and use stoichiometry.
The balanced equation for the reaction is:
K2SO4 + Sr(NO3)2 -> SrSO4 + 2KNO3
To find the molar mass of each compound, you can refer to the periodic table:
- K2SO4: Molar mass = 2 * atomic mass of potassium (K) + atomic mass of sulfur (S) + 4 * atomic mass of oxygen (O)
- Sr(NO3)2: Molar mass = atomic mass of strontium (Sr) + 2 * (atomic mass of nitrogen (N) + 3 * atomic mass of oxygen (O))
- SrSO4: Molar mass = atomic mass of strontium (Sr) + atomic mass of sulfur (S) + 4 * atomic mass of oxygen (O)
- 2KNO3: Molar mass = 2 * (atomic mass of potassium (K) + atomic mass of nitrogen (N) + 3 * atomic mass of oxygen (O))
Once you have obtained the molar masses, you can use stoichiometry to find the grams of the first reactant (K2SO4) needed to react with 55.8 g of the second reactant (Sr(NO3)2).
Here are the steps to calculate it:
1. Calculate the number of moles of the second reactant (Sr(NO3)2) using its molar mass and the given mass:
moles = mass / molar mass
2. Use the balanced equation to determine the stoichiometric ratio between Sr(NO3)2 and K2SO4:
From the equation, you can see that 1 mole of Sr(NO3)2 reacts with 1 mole of K2SO4.
3. Convert moles of Sr(NO3)2 to moles of K2SO4 using the stoichiometric ratio found in step 2.
4. Convert the moles of K2SO4 to grams using its molar mass:
mass = moles * molar mass
Perform these calculations using the given values and the molar masses mentioned earlier, and you will find the grams of K2SO4 required to react with 55.8 g of Sr(NO3)2.