volume of NH3 at stp?
Answers
Answer:22.4 L
Explanation:
Answer: 22.4 L
Explanation: The fertilizer ammonium sulfate, (NH4)2SO4, is prepared by the reaction of ammonia, NH3, with sulfuric
acid: 2 NH3 (g) + H2SO4 (aq) (NH4)2SO4 (aq)
How many moles of ammonium sulfate are produced if 1.50 moles of ammonia react completely at STP?
Answer: The balanced equation indicates that 2 moles of NH3 react to produce 1 mole of (NH4)2SO4.
1.50 moles NH3 x
1 mol (NH4)2SO4
2 mol NH3
= 0.750 moles (NH4)2SO4
2) Use the chemical reaction in problem #1 to determine how many moles of ammonia are required to react
with 2.50 moles of sulfuric acid.
Answer: The balanced equation indicates that 2 moles of NH3 react with 1 mole of H2SO4.
2.50 moles H2SO4 x
2 mol NH3
1 mol H2SO4
= 5.00 moles NH3
3) Use the chemical reaction in problem #1 to determine the mass of ammonium sulfate produced if 50.0 L
of ammonia gas react completely at STP.
Answer: The balanced equation indicates that 2 moles of NH3 react to produce 1 mole of (NH4)2SO4.
Because the reaction occurs at STP, 1 mole of NH3 gas occupies 22.4 L.
50.0 L NH3 x
mol NH3
22.4 L NH3
x
1 mol (NH4)2SO4
2 mol NH3
x
132.17 g (NH4)2SO4
mol (NH4)2SO4
= 148 g (NH4)2SO4
4) Use the chemical reaction in problem #1 to determine the volume (in L) of ammonia gas required to
produce 150.0 g of ammonium sulfate at STP.
Answer: The balanced equation indicates that 1 mole of (NH4)2SO4 is produced when 2 moles of NH3
react. Because the reaction occurs at STP, 1 mole of NH3 gas occupies 22.4 L.
150.0 g (NH4)2SO4 x
mol (NH4)2SO4
132.17 g (NH4)2SO4
x
2 mol NH3
1 mol (NH4)2SO4
x
22.4 L NH3
mol NH3
= 50.8 L NH3
5) Propane, C3H8 (g), burns in oxygen to produce carbon dioxide gas and steam. The balanced equation for
the reaction is C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g)
How many moles of oxygen are required to react completely with 5.00 moles of propane?
5.00 moles C3H8 x
5 mol O2
1 mol C3H8
= 25.0 moles C3H8
6) Use the chemical equation in problem #5 to determine how many moles of steam are produced when
10.0 moles of oxygen react completely.
10.0 moles O2 x
4 mol H2O
5 mol O2
= 8.00 moles H2O
7) Use the chemical equation in problem #5 to determine what volume (in L) of carbon dioxide gas is
produced when 5.00 g of propane is burned at STP.
Answer: The balanced equation indicates that 1 mole of C3H8 reacts to produce 3 moles of CO2.
Because the reaction occurs at STP, 1 mole of CO2 gas occupies 22.4L.
5.00 g C3H8 x
mol C3H8
44.11 g C3H8
x
3 mol CO2
1 mol C3H8
x
22.4 L CO2
mol CO2
= 7.62 L CO2
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