What is the percent yield of h2o if 58g h2o are produced by combining 60g o2 and 7g h2?

Answer 1

Balanced equation.

2H2 + O2 >> 2H2O ( now find limiting reactant )

7 grams H2 (1 mole H2/2.016 grams) = 3.472 moles H2

60 grams O2 (1 mole O2/32 grams) = 1.875 moles O2

1.875 moles O2 (2 mole H2/1 mole O2) = 3.75 mole H2 ( checked O2, but I know H2 limits because you do not have 3.75 moles H2, so H2 drives reaction)

3.472 moles H2 (2 mole H2O/2 mole H2)(18.016 grams/1 mole H2O)

= 62.552 grams H2O produced, so;

58 grams/62.552 times 100

= 92.7% yield of H2O, call it 93%

Answer 2

Oh, dude, percent yield is like the amount of actual stuff you get compared to what you could have gotten if everything was perfect. So, if you produced 58g of H2O but could have gotten 65g if everything went smoothly, your percent yield would be like 89%. It's basically a way to measure how good you are at chemistry without actually having to do math.

Answer 3

The first step is to calculate the theoretical yield of H2O using stoichiometry. First, determine the limiting reactant, which is H2 in this case. Then, use the balanced chemical equation to find the theoretical yield of H2O. Next, calculate the percent yield by dividing the actual yield (58g) by the theoretical yield and multiplying by 100. The percent yield of H2O is approximately 86%.

Answer 4

Theoretical yield of H2O is 65g (from the limiting reactant H2). Percent yield = (actual yield / theoretical yield) * 100. Therefore, percent yield = (58g / 65g) * 100 = 89.2%.

Answer 5

98