In the following reaction, eight moles of sodium hydroxide is broken down into four moles of sodium oxide and four moles of water.
What is the percent error if your experiment yields 195 grams of sodium oxide?
2NaOH→Na2O+H2O

For the given chemical reaction, 8 moles of the reactant should produce 4 moles of [tex]Na_{2}O[/tex]. However, 195 g of [tex]Na_{2}O[/tex] was produced instead. The molar mass of [tex]Na_{2}O[/tex] is 61.9789 g/mol.

Thus, the moles of [tex]Na_{2}O[/tex] produced = 195/61.9789 = 3.1462 moles

The percent error = [(Actual -Experiment)/Actual]*100%

The percent error = [(4.00 - 3.1462)/4.00]*100% = (0.85376/4.00)*100% = 21.344%

In the following reaction, eight moles of sodium hydroxide is broken down into four moles of sodium oxide and four moles of water. What is the percent error if your experiment yields 195 grams of sodium oxide? 2NaOH→Na2O+H2O

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In the following reaction, eight moles of sodium hydroxide is broken down into four moles of sodium oxide and four moles of water.
What is the percent error if your experiment yields 195 grams of sodium oxide?
2NaOH→Na2O+H2O