how many grams of water are needed to react with 80.0g CaCn2?
Well, let's do some math! The molar mass of CaCn2 is approximately 92.11 g/mol. To figure out how many moles of CaCn2 we have, we divide the given mass by the molar mass: 80.0 g / 92.11 g/mol ≈ 0.868 moles.
Now, for every mole of CaCn2, we need 2 moles of water (H2O) to react. Therefore, we multiply the number of moles of CaCn2 by 2 to find the number of moles of water required: 0.868 moles × 2 = 1.736 moles.
Finally, we can convert the moles of water into grams by multiplying by the molar mass of water, which is approximately 18.015 g/mol: 1.736 moles × 18.015 g/mol ≈ 31.29 grams.
So, you would need approximately 31.29 grams of water to react with 80.0 grams of CaCn2. Hope that brings a smile to your face!
To determine the amount of water needed to react with 80.0g of CaCn2, we need to consider the stoichiometry of the chemical reaction. The balanced chemical equation for the reaction is:
CaCn2 + 2H2O -> CaCO3 + 2NH3
From the balanced equation, we can see that 1 mole of CaCn2 reacts with 2 moles of H2O. We can convert the grams of CaCn2 to moles first, and then use the mole ratio to find the moles of H2O.
The molar mass of CaCn2 is:
40.08 g/mol (Ca) + 26.02 g/mol (C) + 28.01 g/mol (N) = 94.11 g/mol (CaCn2)
Moles of CaCn2 = Mass of CaCn2 / Molar mass of CaCn2
Moles of CaCn2 = 80.0 g / 94.11 g/mol ≈ 0.850 mol
According to the balanced equation, 1 mole of CaCn2 reacts with 2 moles of H2O.
So, the moles of H2O needed = 2 × Moles of CaCn2
Moles of H2O needed = 2 × 0.850 mol = 1.70 mol
Now, to convert the moles of water to grams, we can use the molar mass of water, which is 18.015 g/mol.
Mass of H2O = Moles of H2O × Molar mass of H2O
Mass of H2O = 1.70 mol × 18.015 g/mol ≈ 30.63 g
Therefore, approximately 30.63 grams of water are needed to react with 80.0 grams of CaCn2.
To determine how many grams of water are needed to react with 80.0g of CaCn2, we need to know the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between CaCn2 and water can be written as:
CaCn2 + 2H2O -> Ca(OH)2 + 2HCN
From the balanced equation, we can see that one mole of CaCn2 reacts with two moles of water. So, we need to convert the given mass of CaCn2 to moles, and then use the mole ratio from the balanced equation to find the moles of water required. Finally, we convert the moles of water to grams.
Here's how to calculate it step by step:
1. Determine the molar mass of CaCn2:
Ca: 40.08 g/mol
C: 12.01 g/mol
N: 14.01 g/mol
Adding them up: (40.08 g/mol) + (2 × 12.01 g/mol) + (2 × 14.01 g/mol) = 80.08 g/mol
2. Convert the given mass of CaCn2 to moles:
Moles of CaCn2 = (given mass of CaCn2) / (molar mass of CaCn2)
Moles of CaCn2 = 80.0 g / 80.08 g/mol = 0.999 moles (rounded to three decimal places)
3. Use the mole ratio from the balanced equation to find the moles of water:
Moles of water = (moles of CaCn2) × (2 moles of water / 1 mole of CaCn2)
Moles of water = 0.999 moles × 2 = 1.998 moles (rounded to three decimal places)
4. Convert the moles of water to grams:
Mass of water = (moles of water) × (molar mass of water)
Assuming the molar mass of water is approximately 18.015 g/mol,
Mass of water = 1.998 moles × 18.015 g/mol = 35.997 g (rounded to three decimal places)
Therefore, approximately 36.0 grams of water are needed to react with 80.0 grams of CaCn2.
Here is a step by step procedure for solving stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html