How many grams of K2SO3 would be required to react completely with 200.0 mL of 0.245 M HCl
K2SO3 + 2HCl --> 2KCl + H2SO3
mols HCl = M x L = ?
Using the coefficients in the balanced equation, convert mols HCl to mols K2SO3.
Now convert mols K2SO3 to grams. g = mols x molar mass = ?
Well, let me do some quick calculations while trying to keep things light-hearted. Are you ready for a dash of chemistry comedy, my friend? Here we go!
To find out how many grams of K2SO3 are needed, we need to use a bit of stoichiometry.
First, we need to know the balanced chemical equation for the reaction between K2SO3 and HCl. Unfortunately, I don't have that information ready, so I'll have to skip my chemistry comedy routine.
Instead, I'll tell you the general strategy. You need to find the molar ratio between K2SO3 and HCl. Then you can use the equation:
moles of K2SO3 = (volume of HCl in L) × (molarity of HCl) × (molar ratio between K2SO3 and HCl)
Once you have the moles of K2SO3, you can convert that into grams using its molar mass. Now go ahead and do some math, my friend! Just make sure to use appropriate units and consult a periodic table for molar masses.
To determine the amount of K2SO3 needed to react completely with 200.0 mL of 0.245 M HCl, we need to use stoichiometry and the balanced chemical equation between K2SO3 and HCl. The balanced equation is as follows:
K2SO3 + 2HCl → 2KCl + H2SO3
From the balanced equation, we can see that 1 mole of K2SO3 reacts with 2 moles of HCl.
First, we need to calculate the number of moles of HCl in 200.0 mL. To do this, we can use the equation:
moles of HCl = concentration (M) × volume (L)
Converting the volume from milliliters to liters:
volume (L) = 200.0 mL / 1000 = 0.200 L
Now, we can calculate the moles of HCl:
moles of HCl = 0.245 M × 0.200 L = 0.049 moles
Since 1 mole of K2SO3 reacts with 2 moles of HCl, we can determine the moles of K2SO3 needed:
moles of K2SO3 = 0.049 moles HCl / 2 = 0.0245 moles
Finally, we can convert moles of K2SO3 to grams using the molar mass of K2SO3, which is 174.26 g/mol:
grams of K2SO3 = 0.0245 moles × 174.26 g/mol = 4.25 grams
Therefore, approximately 4.25 grams of K2SO3 would be required to react completely with 200.0 mL of 0.245 M HCl.
To determine the number of grams of K2SO3 required to react completely with 200.0 mL of 0.245 M HCl, we need to use stoichiometry.
First, let's write a balanced chemical equation for the reaction between K2SO3 and HCl:
2 K2SO3 + 6 HCl -> 3 S + 3 H2O + 4 KCl
From the balanced equation, you can see that 2 moles of K2SO3 react with 6 moles of HCl to produce 4 moles of KCl.
To find the moles of HCl present in the given 200.0 mL of 0.245 M HCl, we can use the formula:
moles = volume (in liters) x concentration (in moles per liter)
Converting the volume from milliliters to liters:
200.0 mL = 200.0 mL * (1 L / 1000 mL) = 0.200 L
Calculating the moles of HCl:
moles of HCl = 0.200 L * 0.245 mol/L = 0.049 mol
According to the stoichiometry of the balanced equation, we know that 6 moles of HCl react with 2 moles of K2SO3.
Setting up a ratio, we can calculate the moles of K2SO3:
moles of K2SO3 = (2 mol K2SO3 / 6 mol HCl) * 0.049 mol HCl = 0.0163 mol K2SO3
Finally, to find the mass of K2SO3, we can use the formula:
mass = moles x molar mass
The molar mass of K2SO3 can be calculated by adding up the atomic masses of its constituent elements:
2(K) + 1(S) + 3(O) = 39.10 g/mol + 32.07 g/mol + 3(16.00 g/mol) = 158.27 g/mol
Calculating the mass of K2SO3:
mass of K2SO3 = 0.0163 mol K2SO3 * 158.27 g/mol = 2.57 g
Therefore, approximately 2.57 grams of K2SO3 would be required to react completely with 200.0 mL of 0.245 M HCl.