____NaNO3 + ___PbO --> ___Pb(NO3)2 + ___Na[2]O
To balace the eqaution, you need to have the same number of atoms for each element on both the reactant (left) and product (right) side.
To start off, you wanna know the number of atoms in each element on both sides, so take it apart:
[reactants] [product]
Na- 1 Na- 2
N- 1 N- 2(it's 2 because the the subscript [2] is outside of the parenthesis)
O- 4 O- 7 (same reason as above)
Pb- 1 Pb- 1
Na is not balanced out, so add a coefficient to make it the same on both sides.In this case, multiply by 2:
2NaNO3
Now Na is balanced, but the N and O are also effected by this, so they also have to be multiplied by 2 and they become:
Na- 2 Na- 2
N- 2 N- 2 (it balanced out)
O- 7 (coefficient times subscript, plus lone O) O- 7 (balanced out)
Pb was already balanced so no need to mess with it, just put a 1 where needed (it doesn't change anything).
Now to put it back together, it will look like this:
2NaNO3 + 1PbO --> 1Pb(NO3)2 + 1Na[2]O
Hey there!
The answer as well as the explanation is in the image attached. Let me know if there's anything you're unable to see.
Hope this helps!
Explanation:
no.A (He) I think.............
They cause most of their damage when they reach land
Answer:
Percentage yield = 85.2%
Explanation:
Given data:
Mass of Mg = 21.3 g
Actual yield of MgO = 30.2 g
Percentage yield = ?
Solution:
Chemical equation:
2Mg + O₂ → 2MgO
Number of moles of Mg = mass/molar mass
Number of moles of Mg = 21.3 g / 24.3 g/mol
Number of moles of Mg = 0.88 mol
Now we will compare the moles of MgO with Mg.
Mg : MgO
2 : 2
0.88 : 0.88
Mass of MgO:
Mass of MgO= moles × molar mass
Mass of MgO= 0.88 mol × 40.3g/mol
Mass of MgO = 35.46 g
Actual yield of MgO = 30.2 g
Percentage yield:
Percentage yield = Actual yield/theoretical yield × 100
Percentage yield = 30.2 g/ 35.46 g × 100
Percentage yield = 85.2%
