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rizzlepuff8 rizzlepuff8
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12 years ago
If 38 ml of 1.8 M of sulfuric acid is mixed with 5.21 g of zinc and allowed to react together.

Question is above. Please show me steps thanks
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#1
12 years ago
The limiting reagent is the one that produces the least amount of product.

Your reaction is Zn + H2SO4 Rightwards Arrow ZnSO4 + H2.

One mole of Zn produces 1 mole of ZnSO4. You have 5.21 g / 65.30 = 0.0797 mol of Zn, which produces 0.0797 mol of Zn.

One mole of H2SO4 produces one mole of ZnSO4. You have 0.038L * 1.8 moles / L = 0.0684 mol of H2SO4.

Thus H2SO4 is the limiting reagent, and your theoretical yield of ZnSO4 is 0.0684 mol * 161.47 g /mol = 11.0 g ZnSO4.
wrote...
#2
12 years ago
Find the mass of sulfuric acid. Isolate mass in the molarity formula.

M = n solute / V solution (L)
M = (mass / molar mass) / V solution (L)
M(V solution L) = mass / molar mass
M(V solution L)(molar mass) = mass

Solve for the mass.
M(V solution L)(molar mass) = mass
1.8(0.038)(98.08) = mass
6.708672 g = mass

We can now move on to determining the limiting reactant. First, we set out the chemical equation, and balance it if necessary.

H?SO? + Zn ---> ZnSO? + H?

For sulfuric acid:
reactant = moles / coeffiicient
reactant = (mass / molar mass) / coefficient
reactant = (6.708672 g / 98.08 g/mol) / 1
reactant = 0.0684

For zinc:
reactant = moles / coeffiicient
reactant = (mass / molar mass) / coefficient
reactant = (5.21 g / 65.39 g/mol) / 1
reactant = 0.079675791

Since 0.0684 < 0.079675791, sulfuric acid is our limiting reactant.

Calculate the theoretical yield.
t.yield = 6.708672 g (1 mol/98.08 g/mol) (1 mol/1 mol) (161.45/1 mol)
t.yield = 11.04318 or 11.0 g of ZnSO? (Answer)

Hope this helps!
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