Convert the 48.7 grams into moles and then use the balanced equation to convert to moles of hydrogen gas. Then divide by avogadro's number<span />
The reaction is properly written as
Mg₃N₂ (s) + 3 H₂O (l) --> 2 NH₃<span> (g) + 3 MgO (s)
Molar mass of Mg</span>₃N₂ = 100.95 g/mol
Molar mass of H₂O = 18 g/mol
Molar mass of MgO = 40.3 g/mol
Moles Mg₃N₂: 3.82/100.95 = 0.0378
Moles H₂O: 7.73/18 = 0.429
Theo H₂O required for available Mg₃N₂: 0.0378*3/1 = 0.1134 mol
Hence, the limiting reactant is Mg₃N₂.
Thus,
Theoretical Yield = 0.0378 mol Mg₃N₂ * 3 mol MgO/Mg₃N₂ * 40.3 g/mol
Theo Yield = 4.57 g
Percent Yield = Actual Yield/Theo Yield * 100
Percent Yield = 3.60 g/4.57 g * 100 =<em> 78.77%</em>
Answer:
Pb^+7
Explanation:
I placed the carrot there to signify the +7 is supposed to be small. It is plus seven because seven nuetrons are added
The two half-reactions are...
Ag→Ag+
and...
NO3→NO
Let's start by balancing the first half-reaction...
Ag→Ag+
The amounts are already balanced; 1:1. The oxygens are balanced. So all that's left is to balance the charge...
Ag→Ag++e−
Now let's do the other equation... Amounts of nitrogen are balanced, so we first need to balance the oxygens...
NO3→NO
4H++NO3→NO+2H2O
Next, we need to balance charge...
4e−+4H++NO3→NO+2H2O
Now let's go ahead and rewrite each half-reaction after being balanced by themselves...
Ag→Ag++e−
4e−+4H++NO3→NO+2H2O
Now we need to multiply by some factor to get the electrons to cancel out. In this case, that factor is 4, which needs to be applied to the top half-reaction...
4(Ag→Ag++e−)=4Ag→4Ag++4e−
Then we combine this half-reaction with the second one above to get...
4Ag+4H++NO3→4Ag++NO+2H2O
