<span>Mg + O2 > MgO. In reactant side, 2 O atoms and 1 Mg are present. In product side, 1 Mg and O atoms are present. Put 2 in product side to balance O atoms and 2 at Mg in reactant side to balance Mg atoms. Therefore the balanced equation becomes, 2Mg + O2 ----> 2MgO. Hope it helps.</span>
Answer:
53.6 g of N₂H₄
Explanation:
The begining is in the reaction:
N₂(g) + 2H₂(g) → N₂H₄(l)
We determine the moles of each reactant:
59.20 g / 28.01 g/mol = 2.11 moles of nitrogen
6.750 g / 2.016 g/mol = 3.35 moles of H₂
1 mol of N₂ react to 2 moles of H₂
Our 2.11 moles of N₂ may react to (2.11 . 2) /1 = 4.22 moles of H₂, but we only have 3.35 moles. The hydrogen is the limiting reactant.
2 moles of H₂ produce at 100 % yield, 1 mol of hydrazine
Then, 3.35 moles, may produce (3.35 . 1)/2 = 1.67 moles of N₂H₄
Let's convert the moles to mass:
1.67 mol . 32.05 g/mol = 53.6 g
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.
Answer:
10.85 g of water
Explanation:
First we write the balanced chemical equation
Then we calculate the number of moles of nitric acid produced
n(HNO3) = 
According to the balanced equation, water needed in moles is always half the number of moles of HNO3 produced. So since we will produce 1.2044 mol of HNO3, we will need 0.6022 mol of water. Now to calculate what mass that is:
mass(water)=number of moles*molar mass=0.6022mol*18.02g/mol=10.85g
