A camping lantern uses the reaction of calcium carbide, CaC2(s),and water to produce acetylene gas, C2H2(g), and calcium hydroxide, Ca(OH)2(s). How many grams of water are required to produce 1.55 moles of acetylene gas?
To determine the number of grams of water required to produce 1.55 moles of acetylene gas, we need to use the balanced chemical equation for the reaction between calcium carbide and water.
The balanced chemical equation for the reaction is:
CaC2(s) + 2 H2O(l) -> C2H2(g) + Ca(OH)2(s)
According to the equation, we can see that 1 mole of calcium carbide reacts with 2 moles of water to produce 1 mole of acetylene gas. Therefore, the molar ratio between water and acetylene gas is 2:1.
To find the number of moles of water required, we can use the molar ratio:
1.55 moles C2H2 x (2 moles H2O / 1 mole C2H2) = 3.10 moles H2O
Now, to convert the moles of water to grams, we need to use the molar mass of water. The molar mass of water (H2O) is approximately 18 g/mol.
Therefore, the mass of water required is:
3.10 moles H2O x 18 g/mol = 55.8 g of water
So, approximately 55.8 grams of water are required to produce 1.55 moles of acetylene gas.