The answer is 57.14%.
First we need to calculate molar mass of <span>NaHCO3. Molar mass is mass of 1 mole of a substance. It is the sum of relative atomic masses, which are masses of atoms of the elements.
Relative atomic mass of Na is 22.99 g
</span><span>Relative atomic mass of H is 1 g
</span><span>Relative atomic mass of C is 12.01 g
</span><span>Relative atomic mass of O is 16 g.
</span>
Molar mass of <span>NaHCO3 is:
22.99 g + 1 g + 12.01 g + 3 </span>· <span>16 g = 84 g
Now, mass of oxygen in </span><span>NaHCO3 is:
3 </span>· 16 g = 48 g
mass percent of oxygen in <span>NaHCO3:
48 g </span>÷ 84 g · 100% = 57.14%
Therefore, <span>the mass percent of oxygen in sodium bicarbonate is 57.14%.</span>
Explanation:
To solve this problem, follow these steps;
- Obtain a balanced equation of the reaction and familiarize with the reactants and products.
- Find the number of moles of the reacting species since they are the known matter in terms of quantity.
- From the number of moles, determine the limiting reactant.
- The limiting reactant is the one given in short supply.
- Such reactant determines the extent of the reaction.
- Compare the moles of this specie to that of the products using the balanced equation.
- Obtain the mole of the desired product and find the mass or desired quantity.
- simply work from the known specie to the unknown
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Number of moles brainly.com/question/13064292
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