Answer:
![\boxed {\boxed {\sf 76.2 \% }}](https://img.qammunity.org/2021/formulas/chemistry/high-school/k1mek9l5vams3qq9gso5kd6zinodkm5pco.png)
Step-by-step explanation:
A percent yield helps measure how successful or precise a reaction was. It is ratio of measured yield to theoretical yield as a percent. The formula is:
![percent \ yield=(actual \ yield)/(theoretical \ yield ) *100](https://img.qammunity.org/2021/formulas/chemistry/high-school/gc0el73cgjxk9hz3wkvh4kxi77kg948wkn.png)
For this reaction, the measured/actual yield was 278 grams, while the theoretical yield was 365 grams.
![actual = 278 \ g \\\theoretical= 365 \ g](https://img.qammunity.org/2021/formulas/chemistry/high-school/5gz6nd865qba5kf6kpjb71htgifibufsh6.png)
Substitute the values into the formula.
![percent \ yield =(278 \ g)/(365 \ g) *100](https://img.qammunity.org/2021/formulas/chemistry/high-school/ke4degz0zb4v9ze885o0doszhvj4z62khx.png)
Divide first. The grams (g) will cancel out.
- 278 g/365 g= 278/365=0.7616438356
![percent \ yield =0.76164383561*100](https://img.qammunity.org/2021/formulas/chemistry/high-school/59ltty7du5u7le5ixiph4tweestyy260py.png)
Multiply.
![percent \ yield=76.1643835616](https://img.qammunity.org/2021/formulas/chemistry/high-school/gvvdlf2zdps2mopzfbto3a7m5ev4cpvhpn.png)
The original measurements both had 3 significant figures, so we should round our percent yield to 3 sig figs. In this case, that is the tenths place.
The 6 in the tenth place tells us to round the 1 up to a 2.
![percent \ yield \approx 76.2](https://img.qammunity.org/2021/formulas/chemistry/high-school/aurk2ljd62fe2ulpe391tplixm761t8o0h.png)
The percent yield is about 76.2 %