Answer:
Sr(OH)₂(aq) + 2 HNO₃(aq) ⇒ Sr(NO₃)₂(aq) + 2 H₂O(l)
Explanation:
Let's consider the unbalanced molecular equation between strontium hydroxide and nitric acid solutions. This is a neutralization reaction, since an acid reacts with a base to form a salt and water.
Sr(OH)₂(aq) + HNO₃(aq) ⇒ Sr(NO₃)₂(aq) + H₂O(l)
We have 2 N atoms on the right side, so we have to multiply HNO₃ by 2.
Sr(OH)₂(aq) + 2 HNO₃(aq) ⇒ Sr(NO₃)₂(aq) + H₂O(l)
Finally, to get the balanced equation we will multiply H₂O by 2.
Sr(OH)₂(aq) + 2 HNO₃(aq) ⇒ Sr(NO₃)₂(aq) + 2 H₂O(l)
Since there are given data of two reagents, we need to obtain which of reagents are surplus.
0.3618 mol C4H4 - x mol O2
1 mol C4H4 - 5 mol O2
that means that there are too much O2 gas (by 0.009 mol)
so we continue counting using C4H4 data
0.3618 mol C4H4 - x mol H2O
1 mol C4H4 - 2 mol H2O
Mw(H2O)=2×1+16=18 g/mol
n=m/Mw
m=n×Mw
answer:13.0248 grams of H2O
This is a neutralisation reaction, so we know that the products formed are salt and water.
First, we balance the chemical equation:
KOH + HNO3 ----> KNO3 + H2O.
After that, we split the compounds into their respective charges.
K+ + OH- + H+ + NO3- ----------> K+ + NO3- + H2O*
(*note that liquid cannpt be splitted.)
After that we reduce the chemical equation by removing the common ones.
OH- + H+ --------> H2O
Tada.