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Chemical Reaction Lab Report

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Molar Mass of NaHCO3 = 84.098 g/mol Moles in 0.10g = 0.10/84.098 = 0.00119 mol Moles in 0.20g = 0.20/84.098 = 0.00238 mol Moles in 0.30g = 0.30/84.098 = 0.00357 mol Moles in 0.40g = 0.40/84.098 = 0.00476 mol Moles in 0.50g = 0.50/84.098 = 0.00595 mol Moles in 0.90g = 0.90/84.098 = 0.0107 mol Molar mass of CH3COOH = 60.052 g/mol Moles in 5g = 5/60.052 = 0.0833 mol Amount of grams in 5mL of CH3COOH = 5mL* 1g/1mL = 5g CH3COOH The given balanced equation for the chemical reaction shows a 1:1 ratio: NaHCO3 would be the limiting reactant, as CH3COOH has excess and is limited by the amount of NaHCO3 present – once the NaHCO3 is depleted, the reaction stops. To calculate the mass of CO2 produced by taking the number of moles of CO2 multiplied

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