when 31g of copper(2)carbonate is used how many grams of co2 form ?
Hard to say, since you don't specify the reaction.
However, you could have something like
CuCO3 + ... -> Cu?? + CO2 + H2O
That would release 1 mole of CO2 for every mole of CuCO3.
Now just convert 31g CuCO3 to moles, and convert that back to grams of CO2
If that's not your reaction, then write the balanced equation and compare moles of CuCO3 and CO2
Well, if I put on my "wise cracking" hat, I would say that when 31g of copper(2)carbonate is used, you might expect the formation of enough CO2 to fill a clown car! But let's put the silliness aside and get to the calculation.
To figure out how many grams of CO2 are formed, we need to know the molar mass of copper(2)carbonate. Copper(2)carbonate consists of one copper atom (Cu), one carbon atom (C), and three oxygen atoms (O).
So, the molar mass of copper(2)carbonate is:
Cu: 63.55 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
Adding it all up, we get:
63.55 g/mol + 12.01 g/mol + (3 × 16.00 g/mol) = 123.55 g/mol
Now, we can use stoichiometry to find the number of moles of CO2 produced. According to the balanced equation:
2 CuCO3 → 2 CuO + 4 CO2
We have 31g of CuCO3. So, the number of moles of CuCO3 is:
31 g / 123.55 g/mol = 0.251 mol
From the balanced equation, we see that for every 2 moles of CuCO3, 4 moles of CO2 are produced. Therefore, the number of moles of CO2 is twice the number of moles of CuCO3:
0.251 mol × 2 = 0.502 mol
Finally, to find the mass of CO2 formed, we multiply the number of moles by the molar mass of CO2:
0.502 mol × 44.01 g/mol = 22.11 g
So, when 31g of copper(2)carbonate is used, approximately 22.11 grams of CO2 should form.
To determine the number of grams of CO2 formed when 31g of copper(II) carbonate (CuCO3) is used, we need to calculate the molar mass of CuCO3 and then apply stoichiometry.
1. Calculate the molar mass of CuCO3:
- The molar mass of copper (Cu) is 63.55 g/mol.
- The molar mass of carbon (C) is 12.011 g/mol.
- The molar mass of oxygen (O) is 16.00 g/mol.
Molar mass of CuCO3 = (Cu atomic mass) + (C atomic mass) + (3 x O atomic mass)
= (63.55) + (12.011) + (3 x 16.00)
= 123.55 g/mol
2. Calculate the number of moles of CuCO3 used:
Moles of CuCO3 = Mass of CuCO3 / Molar mass of CuCO3
= 31 g / 123.55 g/mol
≈ 0.251 moles (rounded to three decimal places)
3. Apply stoichiometry using the balanced chemical equation for the decomposition of CuCO3:
CuCO3(s) --> CuO(s) + CO2(g)
From the equation, we see that one mole of CuCO3 produces one mole of CO2. Therefore, the number of moles of CO2 formed is also 0.251 moles.
4. Calculate the mass of CO2 formed:
Mass of CO2 = Moles of CO2 x Molar mass of CO2
= 0.251 moles x 44.01 g/mol (molar mass of CO2)
≈ 11.05 g (rounded to two decimal places)
Therefore, approximately 11.05 grams of CO2 will form when 31 grams of copper(II) carbonate is used.
12
I assume this is a decomposition reaction.
CuCO3 ==> CuO + CO2
mols CuCO3 = grams/molar mass = ?
1 mol CuCO3 produces 1 mols CO2.
grams CO2 = mols CO2 x molr mass CO2 = ?