Acetylene gas c2h2 reaction cac2+2h2o. C2h2+Ca(oh)2 how many moles of cac2 needed to prepare 10.0 g c2h2 please show work and answer soon final question on big test review
Indicate your specific subject in the "School Subject" box, so those with expertise in the area will respond to the question.
To find the number of moles of CaC2 needed to prepare 10.0 g of C2H2, we'll first start by finding the molar mass of C2H2.
The molar mass of C2H2 (acetylene) is calculated as follows:
C: 2 atoms x atomic mass of carbon (C) = 2 x 12.01 g/mol = 24.02 g/mol
H: 2 atoms x atomic mass of hydrogen (H) = 2 x 1.01 g/mol = 2.02 g/mol
Total molar mass of C2H2 = 24.02 g/mol + 2.02 g/mol = 26.04 g/mol
Now, using the balanced chemical equation:
2 CaC2 + 2 H2O → C2H2 + Ca(OH)2
We can see that for every 2 moles of CaC2, we produce 1 mole of C2H2.
To convert the mass of C2H2 into moles, we need to use the equation:
n = mass / molar mass
n (moles) = 10.0 g / 26.04 g/mol
n ≈ 0.384 moles (rounded to three decimal places)
Since the mole ratio between CaC2 and C2H2 is 2:1, we will need twice as many moles of CaC2. Thus, the number of moles of CaC2 required is:
2 x 0.384 moles = 0.768 moles
Therefore, to prepare 10.0 g of C2H2, you would need approximately 0.768 moles of CaC2.