Underground miners formerly used headlamps in which calcium carbide (CaC2) was reacted with water to form acetylene (C2H2) gas as shown in the following reaction:
CaC2(s) + H2O (l) ---> C2H2 (g) +Ca(OH2)(s)
The acetylene produced was then burned to produce a very bright light. Starting with 84.0g of calcium carbide and excess water, what volume of acetylene will be produced at 300 K and 1520 torr?
Now I balanced the reaction so that it becomes this:
CaC2(s)+2H2O(l)-->C2H2(g)+Ca(OH2)(s)
and 1520 torr equals about 2 atm but I'm not really sure where else to go from here.
I have corrected your equation.
CaC2(s)+2H2O(l)-->C2H2(g)+Ca(OH)2(s)
mols CaC2 = 84g/molar mass = ?
Using the coefficients in the balanced equation, convert mols CaC2 to mols C2H2.
Then convert mols C2H2 to volume at the conditions listed using PV = nRT and solve for V in liters. For P you will need to convert 1520 torr to atm.
To solve this problem, you will need to use the concept of stoichiometry and the ideal gas law.
1. Start by converting the given mass of calcium carbide (CaC2) into moles. The molar mass of CaC2 can be found by adding the atomic masses of calcium (Ca = 40.08 g/mol) and carbon (C = 12.01 g/mol), multiplied by 2 because there are two carbon atoms in CaC2. So, the molar mass of CaC2 is 40.08 + (12.01 * 2) = 64.10 g/mol.
Moles of CaC2 = Mass of CaC2 / Molar mass of CaC2
= 84.0 g / 64.10 g/mol
2. Now, determine the moles of acetylene (C2H2) that can be produced by comparing the coefficients in the balanced chemical equation. From the balanced equation, 1 mole of CaC2 produces 1 mole of C2H2.
Moles of C2H2 = Moles of CaC2
3. Next, we need to calculate the volume of acetylene gas produced. To do that, we will use the ideal gas law:
PV = nRT
P = pressure (1520 torr, which is approximately equal to 2 atm)
V = volume (unknown)
n = moles of C2H2 (calculated in step 2)
R = ideal gas constant (0.0821 L·atm/(K·mol))
T = temperature (300 K)
Substitute the known values into the equation:
(2 atm)(V) = (moles of C2H2)(0.0821 L·atm/(K·mol))(300 K)
4. Rearrange the equation to solve for V (volume):
V = (moles of C2H2)(0.0821 L·atm/(K·mol))(300 K) / (2 atm)
5. Plug in the calculated value for moles of C2H2 from step 2 and evaluate to find the volume of acetylene gas produced.
Remember to pay attention to significant figures in your calculations and round your final answer to an appropriate number of significant figures.