How many grams of caco3 will be formed by passing excess co2 gas through 100ml of 0.25M ca(H)2 solution?
my experience is formate calcium will form (Ca(HCO2)2 ) will formin that reaction. So I will play along and try to get CaCO3 (s) instead.
Wondering if you meant Ca(OH)2 solution.....
Assuming you did mean that
Ca(OH)2 + CO2 >>>Ca(CO3) + H2O
so moles, it is one to one on Ca(OH)2 vs Ca(CO3)
moles Ca(OH)2: .1*.25=.025 moles, so you form .025 moles calcum carbonate, which in grams is .025*(100)=2,5 grams
check my calculations
Well, to find out how many grams of CaCO3 will be formed, we need to do a little chemical clownery! First, let's find out how many moles of Ca(H)2 we have in 100 ml of 0.25M solution.
Given that the Molarity (M) is moles per liter, we need to convert 100 ml to liters. That's 0.1 L.
So, using the formula M = moles/L, we can rearrange it to solve for moles: moles = M * L.
Plugging in the values: moles = 0.25 mol/L * 0.1 L, we get 0.025 moles of Ca(H)2.
Now, let's look at the balanced chemical equation for the reaction between Ca(H)2 and CO2:
Ca(H)2 + CO2 -> CaCO3 + H2O
The stoichiometry tells us that one mole of Ca(H)2 reacts with one mole of CO2 to form one mole of CaCO3.
So, since we have 0.025 moles of Ca(H)2, we would need 0.025 moles of CO2 to react completely with it.
Now, to find the grams of CaCO3 formed, we need to use the molar mass of CaCO3.
The molar mass of CaCO3 is approximately 100.1 g/mol.
So, to find the grams of CaCO3, we multiply the moles of CaCO3 with the molar mass:
grams = moles * molar mass = 0.025 mol * 100.1 g/mol = 2.5025 grams.
Thus, approximately 2.5025 grams of CaCO3 will be formed when excess CO2 gas is passed through 100 ml of 0.25M Ca(H)2 solution.
And that's how chemistry can make you weigh your options, and your grams! 😄🔬
To determine the grams of CaCO3 formed, we will need to use stoichiometry based on the balanced chemical equation for the reaction between Ca(H)2 and CO2:
Ca(H)2 + CO2 → CaCO3 + H2O
From the balanced equation, we can see that for every 1 mol of Ca(H)2, 1 mol of CaCO3 is formed.
To find the number of moles of Ca(H)2 in the solution, we can use the formula:
moles = concentration (M) × volume (L)
Given:
Concentration of Ca(H)2 solution = 0.25 M
Volume of Ca(H)2 solution = 100 ml = 0.1 L
moles of Ca(H)2 = 0.25 M × 0.1 L = 0.025 mol
Since the reaction is in excess CO2, all of the Ca(H)2 will be consumed, and the same amount of CaCO3 will be formed.
Now, let's calculate the molar mass of CaCO3:
Mass of Ca = 40.08 g/mol
Mass of C = 12.01 g/mol
Mass of 3 O = 3 × 16.00 g/mol = 48.00 g/mol
Molar mass of CaCO3 = 40.08 g/mol + 12.01 g/mol + 48.00 g/mol = 100.09 g/mol
Finally, we can calculate the grams of CaCO3 formed using the moles of CaCO3 and the molar mass of CaCO3:
grams of CaCO3 = moles of CaCO3 × molar mass of CaCO3
= 0.025 mol × 100.09 g/mol
= 2.5025 g
Therefore, passing excess CO2 gas through 100 ml of 0.25 M Ca(H)2 solution will result in the formation of 2.5025 grams of CaCO3.
To determine the number of grams of CaCO3 formed, we need to start by calculating the number of moles of Ca(H)2 in the solution.
Step 1: Calculate the number of moles of Ca(H)2 in the solution.
Given:
Volume of Ca(H)2 solution (V) = 100 mL = 0.1 L
Concentration of Ca(H)2 solution (C) = 0.25 M
We can use the formula:
moles (n) = concentration (C) × volume (V)
n = C × V
= 0.25 mol/L × 0.1 L
= 0.025 mol
Step 2: Write the balanced chemical equation for the reaction between Ca(H)2 and CO2.
Ca(H)2 + CO2 -> CaCO3 + H2O
From the equation, it is clear that 1 mole of Ca(H)2 reacts with 1 mole of CO2 to produce 1 mole of CaCO3.
Step 3: Calculate the number of moles of CaCO3 formed.
Since Ca(H)2 is in excess, the number of moles of CaCO3 formed will be equal to the number of moles of CO2 reacted.
Therefore, the number of moles of CaCO3 formed = number of moles of CO2
Step 4: Convert moles of CaCO3 to grams.
To convert moles of CaCO3 to grams, we need to multiply the number of moles by its molar mass, which can be calculated from the periodic table.
Molar mass of CaCO3 = (1 × atomic mass of Ca) + (1 × atomic mass of C) + (3 × atomic mass of O)
Molar mass of CaCO3 ≈ 40.08 g/mol + 12.01 g/mol + (3 × 16.00 g/mol)
≈ 100.09 g/mol
Now, we can calculate the grams of CaCO3 formed:
Mass of CaCO3 = number of moles of CaCO3 × molar mass of CaCO3
= 0.025 mol × 100.09 g/mol
≈ 2.50225 g
Therefore, approximately 2.50225 grams of CaCO3 will be formed by passing excess CO2 gas through 100 mL of 0.25 M Ca(H)2 solution.