Hey there!:
Write the molecular equation for the reaction of MgSO4 with Pb(NO3)2 :
MgSO4(aq) + Pb(NO3)2(aq) ---> Mg(NO3)2(aq) + PbSO4(s)
Write the total ionic equation for the reaction :
Mg²⁺ (aq) + SO₄⁻² (aq) + Pb²⁺ (aq) + 2 NO₃⁻¹ (aq) + PbSO₄(s)
Therefore:
Cancel the spectator ions on both sides:
Pb²⁺ (aq) + SO₄⁻² (aq) ---> PbSO4(s)
Hope that helps!
Answer:
I think it's answer is P4O6
I hope it's helpful for you...
<span>a. Use PV = nRT and solve for n = number of mols O2.
mols NO = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols O2 to mols NO2. Do the same for mols NO to mols NO2. It is likely that the two values will not be the same which means one is wrong; the correct value in LR (limiting reagent) problems is ALWAYS the smaller value and the reagent producing that value is the LR.
b.
Using the smaller value for mols NO2 from part a, substitute for n in PV = nRT, use the conditions listed in part b, and solve for V in liters. This will give you the theoretical yield (YY)in liters. The actual yield at these same conditions (AY) is 84.8 L.
</span>and % will be 60%.
Answer:
2 mol H₂O
Explanation:
With the reaction,
- 2H₂(g) + O₂(g) → 2 H₂O(g)
1.55 moles of O₂ would react completely with ( 2*1.55 ) 3.1 moles of H₂. There are not as many moles of H₂, thus H₂ is the limiting reactant.
Now we <u>calculate the moles of H₂O produced</u>, <em>starting from the moles of limiting reactant</em>:
- 2.00 mol H₂ *
= 2 mol H₂O
