Answer:
NH4Br + AgNO3 —> AgBr + NH4NO3
Explanation:
When ammonium bromide and silver(I) nitrate react, the following are obtained as shown below:
NH4Br(aq) + AgNO3(aq) —>
In solution, NH4Br(aq) and AgNO3(aq) will dissociate as follow:
NH4Br(aq) —> NH4+(aq) + Br-(aq)
AgNO3(aq) —> Ag+(aq) + NO3-(aq)
The double displacement reaction will occur as follow:
NH4+(aq) + Br-(aq) + Ag+(aq) + NO3-(aq) —> Ag+(aq) + Br-(aq) + NH4+(aq) + NO3-(aq)
NH4Br(aq) + AgNO3(aq) —> AgBr(s) + NH4NO3(aq)
Double replacement because H and K are both switching
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Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
yes your are right because water do have a stronger attraction