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Limiting Reagent Lab Report

Satisfactory Essays
Determine limiting reagent

H2SO4 + 2 NaOH → Na2SO4 + 2 H2O

NaOH
H2SO4
Have
1.50g ± 0.67% * 1mol40.0g=0.0366 mol ± 0.67%
0.5 molL *0.02500 L ± 0.24% =0.0013 mol ± 0.24%

NaOH is limiting
0.0366 mol ± 0.67%
0.0366 mol NaOH ±0.67% *1 mol H2SO42 mol NaOH= 0.0183 molH2SO4 ± 0.67%0
H2SO4 is limiting 0.0013 molH2SO4± 0.24% *2 mol NaOH1 mol H2SO4 = 0.0026 mol NaOH ± 0.24%
0.0013 mol ± 0.24%

There is not enough NaOH to complete the reaction therefore H2SO4 is limiting.

Table 2: Summary of data - moles of products

Moles of H2SO4
Moles of NaOH
Moles of NaOH (reacted)
Trial 1
0.0013 mol ± 0.24%
0.0366 mol ± 0.67%

0.0026 mol NaOH
± 0.24%
Trial 2
0.0013 mol ± 0.24%

0.0390 mol ± 0.64%
0.0026 mol NaOH
± 0.24%
Trial 3
0.0013 mol ± 0.24%
…show more content…
Calculate percent error
The theoretical value for the neutralization of NaOH and H2SO4 is -57 kJ *mol-1, according to the CRC Handbook of Chemistry and Physics (W. M. Haynes, 2015)

Percent Error = Experimental Value - Theoretical ValueTheoretical Value* 100

Theoretical Value = - 57 kJ *mol-1
Experimental Value = -80.9 kJ *mol-1

Table 6: Percent Error

Percent Error
Trial 1
40.3%
Trial 2
59.5%
Trial 3
30.9%

The experimental value for the neutralization of NaOH and H2SO4 was closest to the theoretical value, in trial 3, with an percent error of 30.9%. All experimental values calculated were relatively high. This accounts for the heat loss through the coffee cup calorimeter, and the extrapolation of the data to find the maximum
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