You weigh out 10.000 g of a mixture of NaHCO3 and Na2CO3 and add an excess of HCl. After the reaction is over, you evaporate off the water and collect the solid, which is found to weigh 9.400 g. How many grams of the original mixture were NaHCO3?
See your post above.
To find the grams of NaHCO3 in the original mixture, we can use the concept of stoichiometry from the balanced chemical equation.
First, let's write the balanced chemical equation for the reaction between NaHCO3 and HCl:
NaHCO3 + HCl → NaCl + H2O + CO2
From the equation, we can see that 1 mole of NaHCO3 reacts with 1 mole of HCl. The molar mass of NaHCO3 is approximately 84 g/mol.
Now, let's calculate the moles of NaHCO3 that reacted by using the relationship between moles, grams, and molar mass:
10.000 g of the mixture - 9.400 g of the solid collected = 0.6 g of CO2 produced.
To convert grams of CO2 to moles, we need to use the molar mass of CO2, which is approximately 44 g/mol.
Moles of CO2 = (0.6 g CO2) / (44 g/mol CO2) ≈ 0.0136 mol CO2
Since the balanced chemical equation tells us that the moles of NaHCO3 and CO2 are equal, we can conclude that 0.0136 mol of NaHCO3 reacted.
Finally, let's calculate the grams of NaHCO3 in the original mixture:
Grams of NaHCO3 = (0.0136 mol NaHCO3) * (84 g/mol NaHCO3) ≈ 1.14 g
Therefore, approximately 1.14 grams of the original mixture were NaHCO3.