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1. Balance the following: n2 + h2 -> NH3rnrnWhat volume of NH3 at STP is produced if 25.0 g of n2 is reacted with an excess of H2?
Accepted Answer
1. To balance the equation, we need to make sure that the number of atoms of each element is the same on both sides of the equation. In this case, we can balance the equation by multiplying N2 by 3 and H2 by 6:
3N2 + 6H2 → 2NH3
2. Now we can use stoichiometry to determine how much NH3 is produced from 25.0 g of N2. First, we convert the mass of N2 to moles:
moles of N2 = 25.0 g / 28.01 g/mol = 0.892 mol
3. According to the balanced equation, 3 moles of N2 react to produce 2 moles of NH3. So, 0.892 mol of N2 will react to produce:
moles of NH3 = 0.892 mol N2 * (2 mol NH3 / 3 mol N2) = 0.595 mol NH3
4. Finally, we can use the ideal gas law to determine the volume of NH3 produced at STP (standard temperature and pressure, which is 0 °C and 1 atm):
V = nRT / P
where:
V is the volume in liters
n is the number of moles
R is the ideal gas constant (0.0821 Latm/(molK))
T is the temperature in Kelvin (273.15 K at STP)
P is the pressure in atmospheres (1 atm at STP)
Plugging in the values we know:
V = 0.595 mol * 0.0821 Latm/(molK) * 273.15 K / 1 atm
V = 13.3 L
Therefore, 25.0 g of N2 will react with an excess of H2 to produce 13.3 L of NH3 at STP.
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