Balance each one by adding electrons to make the charges on both sides the same:
Sn--> Sn2+ + 2 e-
Ag+ + 1 e- --> Ag
Now, you have to have the same number of electrons in the two half-reactions, so multiply the second one by 2 to get:
2 Ag+ + 2 e- --> 2 Ag
Now, just add the two half reactions together, cancelling anything that's the same on both sides:
2 Ag+ + Sn --> Sn2+ + 2 Ag
And you're done.
The balanced combustion reaction of propane, C₃H₈, is
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
Molar mass of propane: 44 g/mol
Moles of propane = 42 g * (1 mol/44g) = 0.9545 mol propane
Molar mass of oxygen: 32 g/mol
Moles of oxygen = 115 g * (1 mol/32 g) = 3.594 mol oxygen
Moles of oxygen needed to completely react propane:
0.9545 mol propane * (5 mol O₂/1 mol propane) = 4.7725 mol oxygen
Since the available oxygen is only 3.594 moles and propane needs 4.7725 moles, that means oxygen is our limiting reactant. We base the amount of water produced here.
Molar mass of water: 18 g/mol
Mass of water produced = 3.594 mol O₂ * (4 mol H₂O/5 mol O₂) * (18 g/mol)
Mass of water produced = 258.768 grams
Answer: can someone please translate so i can answer.
Explanation:
sorry
Answer:
MgCl2 + 2AgNO3 → 2AgCl + Mg(NO3)2
Explanation:
I'm assuming you want to balance it so...
The first thing I see is that there are two chlorines on the reactant side and one on the product side
Adding a coefficient of 2 would get 2AgCl2
Now there are two silvers on the reactant side, so add a 2 to AgNO3 on the products side. Now they are all balanced.
If that is not what you are looking for let me know!
