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I need help in Chemistry Stoichiometry?
In the equation Na2CO3 + 4C + N2 yields 2 NaCN + 3CO, how would you determine the volume of the carbon monoxide gas when you're given that you have 2.8 grams of N2?
1 Answer
- Anonymous9 years agoFavorite Answer
Na2CO3 + 4C + N2 = 2 NaCN + 3CO
Assuming that N2 is the limiting reagent.
2.8 g of N2 in moles is
n = 2.8 g / 28.01 g/mol = 0.1 mol
Lets look at the equation Na2CO3 + 4C + N2 = 2 NaCN + 3CO
We can see that for every mole of N2 we produced 3 moles of CO
Thats a 1:3 ratio
Therefore 0.1 moles of N2 would produce 3 x 0.1 mol = 0.3 mol of CO
Now lets assume that CO acts as an ideal gas and that we are carrying out this experiment at STP (standard temp and pressure)
n = V / 22.4 L/mol
n x 22.4 L/mol = V
0.3 mol x 22.4 L/mol = 6.72 L
Therefore 6.72 L of CO is produced when 2.8 grams of N2 react.