Anonymous
Anonymous asked in Education & ReferenceHomework Help · 7 years ago

# Chemistry help? Limiting reagent and yield?

I cannot figure this out for the life of me.

Barium sulfate, BaSO4, is made by the following reaction:

Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(aq) + NaNO3(aq)

An experiment began with 75.00 of Ba(NO3)2 and an excess of NaSO4. After collecting and drying the product, 63.45g BaSO4 was obtained. Calculate the theorecitcal yield and percent yield of BaSO4.

Limiting Reagent:

Theoretical Yield:

Percent Yield:

Relevance
• First of all calculate the Mr ( Relative molecular mass ).

Ba(NO3)2

Ba x 1 = 137 x 1 = 137

N x 2 = 14 x 2 = 28

O x 6 = 16 x 6 = 96

137 + 28 + 96 = 261

Similarly

BaSO4

Ba x 1 = 137 x1 = 137

S x 1 = 32 x 1 = 32

O x 4 = 16 x 4 = 64

137 + 32 + 64 = 233

The limiting reagent is barium nitrate, because you have said the sodium sulphate in in excess..

Your reaction equation is slightly wrong.

It I corrected below.

Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(aq) + 2NaNO3(aq)

The molar ratios in the reaction are 1:1::1:2

mol(Ba(NO3)2) = 75.00g / 261 = 0.2874 (Equivalent to '1' in the molar ratios above).

mol(BaSO4) = 0.2874 (Also equivalent to '1' in the molar ratios above).

mass(g)(BaSO4) = 0.2874 x 233 = 66.95 g (This is the 100% theoretical yield).

The percentage yield is 63.45 x 100 / 66.95 = 94,77% (Percent yield).

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