By looking at a property chart and figuring out what will become from certain elements
Answer:
165 g of NaCl are formed in the reaction
Explanation:
2Na + Cl₂ → NaCl
In order to determine the limiting reactant, we convert the mass of each reactant to moles
35 g / 23g/mol = 1.52 moles Na
100 g / 70.9 g/mol = 1.41 moles Cl₂
1 mol of chlorine reacts with 2 moles of Na, so If I have an x value of moles of Cl₂ I would need the double to react.
For 1.41 moles of Cl₂, I need 2.82 moles of Na; therefore my limiting reagent is the Na. Ratio is 2:2. So if I have 2.82 moles of Na I will produce 2.82 moles of NaCl
We convert the moles to mass: 2.82 mol . 58.45 g/1 mol =164.8 g
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The complete balanced chemical reaction for this is:
C6H8O7 + 3 NaHCO3 ---> 3 H2O + 3 CO2 + Na3C6H5O7
First calculate for the number of moles of each reactant.
moles C6H8O7 = 1 g / (192.124 g / mol) = 5.2 * 10^-3 mol
moles NaHCO3 = 1 g / (84.01 g / mol) = 11.9 * 10^-3 mol
A. The ratio of the reactant from the chemical reaction
is 3NaHCO3:1C6H8O7, while the given chemicals are in the ratio of:
11.9 * 10^-3NaHCO3: 5.2 * 10^-3 C6H8O7 =
2.29NaHCO3:1C6H8O7
Therefore this means that there is less amount of NaHCO3
supplied than what is required therefore the limiting reactant is:
NaHCO3
B. We calculate based on the limiting reactant.
mass CO2 = 11.9 * 10^-3 mol NaHCO3 (3 mol CO2/1 mol
NaHCO3) (44.01 g/mol)
mass CO2 = 1.57 g
C. I believe what is asked here is the amount of excess
reactant which remains. The excess reactant is C6H8O7.
mass C6H8O7 left = [5.2 * 10^-3 mol – (11.9 * 10^-3 mol
NaHCO3 (1 mol C6H8O7/ 3 mol NaHCO3))] * (192.124 g / mol)
mass C6H8O7 left = 0.237 g