Answer:
When two forces are the same strength but act in opposite direction, they are called balanced forces. Again, tug-of-war is a perfect example. If the people on each side of the rope are pulling with the same strength, but in the opposite direction, the forces are balanced. The result is no motion.
Explanation:
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Answer:
2HgS + 3O2 → 2HgO + 2SO2
The coefficients are: 2, 3, 2, 2
Explanation:
HgS + O2 → HgO + SO2
The equation can be balance as follow:
Put 3 in front of O2 as shown below:
HgS + 3O2 → HgO + SO2
Now we can see that there are 6 atoms of O on the left side of the equation and a total of 3 atoms on the right side. It can be balance by putting 2 in front of HgO and SO2 as shown below:
HgS + 3O2 → 2HgO + 2SO2
Now we have 2 atoms of both Hg and S on the right side and 1atom each on the left. It can be balance by putting 2 in front of HgS as shown below:
2HgS + 3O2 → 2HgO + 2SO2
Now the equation is balanced.
The coefficients are: 2, 3, 2, 2
The law of conservation of mass(matter) states that matter(mass) can neither be created nor destroyed during a chemical reaction but changes from one form to another. An unbalanced equation suggests that matter has been created or destroyed. While a balanced equation proofs that matter can never be created but changes to different form. This is the more reason we have count the atoms of an element on both side of the equation to see if they are balanced irrespective of the new form they assume in the product
Answer:
See Explanation
Explanation:
The equation of the reaction;
KHSO4(aq) + KOH(aq) -------> K2SO4(aq) + H2O(l)
Number of moles of KHSO4 = 49.6 g/136.169 g/mol = 0.36 moles
Since the reaction is in a mole ratio of 1:1, 0.36 moles of K2SO4 is produced.
Number of moles of KOH = 25.3 g/56.1056 g/mol = 0.45 moles
Since the reaction is 1:1, 0.45 moles of K2SO4 is produced
Hence K2SO4 is the limiting reactant.
Mass of K2SO4 formed = 0.36 moles of K2SO4 * 174.26 g/mol = 62.7 g
So;
1 mole of KHSO4 reacts with 1 mole of KOH
0.36 moles of KHSO4 reacts with 0.36 * 1/1 = 0.36 moles of KOH
Amount of excess KOH = 0.45 moles - 0.36 moles = 0.09 moles
Mass of excess KOH = 0.09 moles * 56.1056 g/mol = 5 g of excess KOH
