Which of the following should occur when placing a certain amount of tetraphosphorus solid in the same container as dihydrogen gas, if #"8.0 mols"# of #"H"_2(g)# are present?
#A)# #"1 mol"# of #"P"_4(s)# reacts.
#B)# #"32 mols"# of #"P"_4(s)# are produced.
#C)# No reaction occurs with #"H"_2(g)# .
#D)# #"16 mols"# of #"PH"_3(g)# is produced.
Solution
C is not possible as we assume the reactants are at a condition where the reaction is possible
D is also not valid as by B
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From the balanced reaction:
#"P"_4(s) + 6"H"_2(g) -> 4"PH"_3(g)# ,we begin with
#"8.0 mols"# of#"H"_2(g)# .Since we have
#"6 mols H"_2# for every#"1 mol P"_4(s)# , we know that
#8.0 cancel("mols H"_2(g)) xx ("1 mol P"_4(s))/(6 cancel("mols H"_2(g)))#
#= ul("1.33 mols P"_4(s))# will react with it. Likewise, since
#"P"_4(s)# is#1:4# with#"PH"_3# , we would produce#ul("5.33 mols")# of#"PH"_3(g)# .
#A)# is wrong, as we have shown it reacts with more than#"1 mol"# of#"P"_4(s)# ... It is the only answer that could potentially have been right, if we started with#"6.0 mols H"_2# ... but we didn't!
#B)# is wrong, because that is far too many mols produced.
#C)# Phosphorus is clearly reactive towards#"H"_2(g)# . It looks like this:
The bond angles (
#60^@# !) are strained, making#"P"_4(s)# quite reactive...
#D)# is wrong, just as#B)# is also wrong. We found only#"5.33 mols"# produced...
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When placing a certain amount of tetraphosphorus solid in the same container as dihydrogen gas, a chemical reaction should occur where the tetraphosphorus solid reacts with the dihydrogen gas to form phosphorus hydride gas. The balanced chemical equation for this reaction is:
P4(s) + 6 H2(g) -> 4 PH3(g)
According to the stoichiometry of the balanced equation, for every 1 mole of tetraphosphorus solid, 6 moles of dihydrogen gas are required. Therefore, if 8.0 moles of dihydrogen gas are present, it would require 8.0/6 = 1.33 moles of tetraphosphorus solid for complete reaction to occur.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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