Acetylene gas combust.how many grams of carbon dioxide and grams of water are produced when 52.0 g of acetylene burns ?
To determine the number of grams of carbon dioxide (CO2) and water (H2O) produced when 52.0 g of acetylene (C2H2) burns, we need to first write a balanced chemical equation for the combustion reaction of acetylene.
The balanced chemical equation for the combustion of acetylene is as follows:
2C2H2 + 5O2 -> 4CO2 + 2H2O
According to this equation, every 2 moles of acetylene produce 4 moles of carbon dioxide and 2 moles of water.
Now, let's calculate the molar mass of acetylene:
Molar mass of C = 12.01 g/mol
Molar mass of H = 1.008 g/mol
Molar mass of acetylene (C2H2) = (2 x molar mass of C) + (2 x molar mass of H)
= (2 x 12.01 g/mol) + (2 x 1.008 g/mol)
= 26.04 g/mol
Next, we'll perform the stoichiometric calculations using the given information:
52.0 g of acetylene x (1 mol / 26.04 g) = 1.996 mol of acetylene
From the balanced equation, we can see that:
2 moles of acetylene produce 4 moles of CO2
2 moles of acetylene produce 2 moles of H2O
Therefore, to calculate the moles of carbon dioxide produced:
1.996 mol of acetylene x (4 mol CO2 / 2 mol acetylene) = 3.992 mol of CO2
And to calculate the moles of water produced:
1.996 mol of acetylene x (2 mol H2O / 2 mol acetylene) = 1.996 mol of H2O
Finally, we can calculate the grams of carbon dioxide and water produced:
Grams of CO2 = moles of CO2 x molar mass of CO2
= 3.992 mol x 44.01 g/mol (molar mass of CO2)
= 175.87 g
Grams of H2O = moles of H2O x molar mass of H2O
= 1.996 mol x 18.015 g/mol (molar mass of H2O)
= 35.93 g
Therefore, when 52.0 g of acetylene burns, approximately 175.87 grams of carbon dioxide and 35.93 grams of water are produced.