If the reaction of 9.87 moles of potassium with excess hydrobromic acid produced a 79.8% yield of hydrogen gas, what was the actual yield of hydrogen gas?

Unbalanced equation: K + HBr “yields”/ KBr + H2

7.88 mol H2

5.74 mol H2

4.94 mol H2

3.94 mol H2

Question

If the reaction of 9.87 moles of potassium with excess hydrobromic acid produced a 79.8% yield of hydrogen gas, what was the actual yield of hydrogen gas?

Unbalanced equation: K + HBr “yields”/ KBr + H2

7.88 mol H2

5.74 mol H2

4.94 mol H2

3.94 mol H2

jfraser · Answer

$$K + HBr \rightarrow KBr + H_2$$First things first, balance the reaction:
If you know the theoretical yield, and the percent yield, you work backwards to find out the actual. The balanced reaction tells you that for every 2mol of K, you'll make 1mol of H2$$2K + 2HBr \rightarrow 2KBr + 1H_2$$

If you start with 9.87mol of K, then you SHOULD produce HALF THAT MUCH H2. THe balanced reaction predicts that for us.

$$9.87mol K * (\frac{1mol H_2}{2mol K})$$Half of 9.87moles is 4.935moles of H2 gas. That yield is 100% of what the reaction SHOULD produce. If only 79.8% of that yield is made, then take 79.8% of the theoretical yield.

79.8% of 4.935 moles is 3.94 moles of H2 produced

anonymous · Answer

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an independent research organization test the product and record the results.

a famous actor or actress endorse the product.

advertisements in well-respected beauty and fashion magazines.

before and after pictures featured in the advertisement.

jfraser · Answer

which one of those is objective rather than subjective?

anonymous · Answer

I really have no idea. i dont understand the quesiton

anonymous · Answer

actually i think its the last answer. before and after pictures.