Asked by Harold
For each of the following precipitation reactions calculate how many grams of the first reactant are necessary to completely react with 55.8 g of the second reactant. K2SO4 + Sr(NO3)2 + SrSO4 + 2KNO3
Answers
Answered by
bobpursley
I think you mean
K2SO4(aq) + Sr(NO3)2(aq) > SrSO4 (s)+ 2KNO3 (aq)
so you need one mole of the first reactant for each mole of the second reactant.
How many moles of strontium nitrtate=55.8/211.6
so you need that number of moles of postassium sulfate, and in grams that is
55.8/211.6 * formulamassK2SO4=
55.8/211.6 *174.3 grams
K2SO4(aq) + Sr(NO3)2(aq) > SrSO4 (s)+ 2KNO3 (aq)
so you need one mole of the first reactant for each mole of the second reactant.
How many moles of strontium nitrtate=55.8/211.6
so you need that number of moles of postassium sulfate, and in grams that is
55.8/211.6 * formulamassK2SO4=
55.8/211.6 *174.3 grams
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.