What volume of hydrogen gas at STP is required for the complete reaction of 28.8 g of nitrogen in the following reaction?
N2(g) + 3 H2(g) → 2 NH3(g)
Could someone please give me a detailed resonse to this so that I can learn how to do it? Thanks!
N2(g) + 3 H2(g) → 2 NH3(g)
Could someone please give me a detailed resonse to this so that I can learn how to do it? Thanks!
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STP calculations are fairly easy. Two small formulas are required in this case. 1) n = m / M,
and 2) n = v / V, where n = moles, v = volume, and V = 22.4 L/mol at STP
So you have your balanced equation. N2(g) + 3 H2(g) → 2 NH3(g)
Notice that it says that for every 1 mol of N2 you have you need to have 3 mol of H2. Right?
Always conveert masses into moles. You have 28.8 g of N2, therefore moles = 28.8 g / 28.2 g/mol
= 2 mol of N2 (rounding off)
So if you have 2 mol of N2 you will need 3 x 2 mol of H2, or 6 mol of H2
Convert that back to grams.
mass = mol x molar mass
6 mol x 2.02 g/mol = 12.12 g
OK?
and 2) n = v / V, where n = moles, v = volume, and V = 22.4 L/mol at STP
So you have your balanced equation. N2(g) + 3 H2(g) → 2 NH3(g)
Notice that it says that for every 1 mol of N2 you have you need to have 3 mol of H2. Right?
Always conveert masses into moles. You have 28.8 g of N2, therefore moles = 28.8 g / 28.2 g/mol
= 2 mol of N2 (rounding off)
So if you have 2 mol of N2 you will need 3 x 2 mol of H2, or 6 mol of H2
Convert that back to grams.
mass = mol x molar mass
6 mol x 2.02 g/mol = 12.12 g
OK?