How many grams of Na2CO3 can be produced from 2.715 g NaHCO3?
Rose, all of these stoichiometry problems are worked the same way.
Step 1. Write the balanced equation.
2NaHCO3 + heat ==> Na2CO3 + H2O + CO2
Step 2. Convert what you have (in this case g NaHCO3) to moles. moles = grams/molar mass
Step 3. Using the coefficients in the balanced equation, convert moles of what you have (in this case moles NaHCO3) to moles of what you want (in this case moles Na2CO3).
moles Na2CO3 = moles NaHCO3 x (2 moles NaHCO3/1 mole Na2CO3) = moles NaHCO3 x (2/1) = ??
Step 4. Now convert moles from step 3 to grams. g = moles x molar mass.
Memorize how to do these four steps You will use them an indeterminate number of times in chemistry.
First, balance the equation:
Na2CO3 + 2HNO2 --> 2NaNO3 + CO2 + H2O
Second, convert 100g of NaNO3 to moles:
100g NaNO3 x 1mol/84.96grams
the grams cancel out, divide 100 by 84.96 to get moles.
Third, use the mole ratio from the equation to solve for Na2CO3 needed:
1.18 mol NaNO3 x 1 mol Na2CO3/2 mol NaNO3
The mole ratio is taken from the balanced equation. Now, cancel out the NaNO3 by dividing 1.18 by 2.
The moles of Na2CO3 needed: .59 moles, using two significant figures.
Did I read the wrong problem? I don't see anything about HNO2.
Well, it seems like we're getting all chemical here! So, to calculate how many grams of Na2CO3 can be produced from 2.715 g of NaHCO3, we need to know the molar ratio between the two compounds.
NaHCO3 (sodium bicarbonate) has a molar mass of approximately 84 g/mol, while Na2CO3 (sodium carbonate) has a molar mass of around 106 g/mol.
So, let's do some math!
First, we need to calculate the number of moles of NaHCO3:
2.715 g NaHCO3 / 84 g/mol NaHCO3 = 0.0323 moles NaHCO3
Now, we need to use the balanced chemical equation to find the mole ratio between NaHCO3 and Na2CO3. Since I'm a Clown Bot and chemistry is not my strongest suit, let's assume the balanced equation is "x NaHCO3 → y Na2CO3."
Based on the balanced equation, we can deduce that for every 2 moles of NaHCO3, we get 1 mole of Na2CO3.
Therefore, the moles of Na2CO3 produced would be:
0.0323 moles NaHCO3 × (1 mole Na2CO3 / 2 moles NaHCO3) = 0.0161 moles Na2CO3
Finally, we need to convert moles of Na2CO3 to grams:
0.0161 moles Na2CO3 × 106 g/mol Na2CO3 = 1.71 g Na2CO3
So, approximately 1.71 grams of Na2CO3 can be produced from 2.715 grams of NaHCO3. Hope that makes you smile like a chemistry clown!
To determine the number of grams of Na2CO3 that can be produced from 2.715 g of NaHCO3, we first need to establish a balanced chemical equation for the reaction involving NaHCO3.
The balanced chemical equation is:
2 NaHCO3 -> Na2CO3 + H2O + CO2
From the balanced equation, we can see that 2 moles of NaHCO3 produce 1 mole of Na2CO3. Thus, we need to convert the mass of NaHCO3 to moles using its molar mass.
The molar mass of NaHCO3 is calculated by adding the atomic masses of the elements:
Na: 22.99 g/mol
H: 1.01 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (since there are three oxygen atoms in NaHCO3)
Molar mass of NaHCO3 = (22.99) + (1.01) + (12.01) + (3 * 16.00) = 84.01 g/mol
Now we can convert the mass of NaHCO3 to moles:
Moles of NaHCO3 = mass of NaHCO3 / molar mass of NaHCO3
= 2.715 g / 84.01 g/mol
≈ 0.0323 mol
Since 2 moles of NaHCO3 produce 1 mole of Na2CO3, we can determine the moles of Na2CO3 produced by dividing the moles of NaHCO3 by 2:
Moles of Na2CO3 = moles of NaHCO3 / 2
≈ 0.0323 mol / 2
≈ 0.0161 mol
Finally, we can convert the moles of Na2CO3 to grams using its molar mass:
Molar mass of Na2CO3 = (22.99 * 2) + (12.01) + (3 * 16.00)
= 105.99 g/mol
Grams of Na2CO3 = Moles of Na2CO3 * molar mass of Na2CO3
≈ 0.0161 mol * 105.99 g/mol
≈ 1.707 g
Therefore, approximately 1.707 grams of Na2CO3 can be produced from 2.715 grams of NaHCO3.