Since you have not included the chemical reaction I will explain you in detail.
1) To determine the limiting agent you need two things:
- the balanced chemical equation
- the amount of every reactant involved as per the chemical equation
2) The work is:
- state the mole ratios of all the reactants: these are the ratios of the coefficientes of the reactans in the balanced chemical equation.
- determine the number of moles of each reactant with this formula:
number of moles = (mass in grams) / (molar mass)
- set the proportion with the two ratios (theoretical moles and actual moles)
- compare which reactant is below than the stated by the theoretical ratio.
3) Example: determine the limiting agent in this reaction if there are 100 grams of each reactant:
i) Chemical equation: H₂ + O₂ → H₂O
ii) Balanced chemical equation: 2H₂ + O₂ → 2H₂O
iii) Theoretical mole ration of the reactants: 2 moles H₂ : 1 mol O₂
iv) Covert 100 g of H₂ into number of moles
n = 100g / 2g/mol = 50 mol of H₂
v) Convert 100 g of O₂ to moles:
n = 100 g / 32 g/mol = 3.125 mol
vi) Actual ratio: 50 mol H₂ / 3.125 mol O₂
vii) Compare the two ratios:
2 mol H₂ / 1 mol O ₂ < 50 mol H₂ / 3.125 mol O₂
Conclusion: the actual ratio of H₂ to O₂ is greater than the theoretical ratio, meaning that the H₂ is in excess respect to the O₂. And that means that O₂ will be consumed completely while some H₂ will remain without react.
Therefore, the O₂ is the limiting reactant in this example.
Answer:
true
Explanation:
in order for the atom to be neutral the number of protons and electrons have to be the same.
Answer:
74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.
Explanation:
The balanced reaction is:
Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:
- Na₂CO₃: 1 mole
- Ca(NO₃)₂: 1 mole
- CaCO₃: 1 mole
- NaNO₃: 2 mole
Being the molar mass of the compounds:
- Na₂CO₃: 106 g/mole
- Ca(NO₃)₂: 164 g/mole
- CaCO₃: 100 g/mole
- NaNO₃: 85 g/mole
then by stoichiometry the following quantities of mass participate in the reaction:
- Na₂CO₃: 1 mole* 106 g/mole= 106 g
- Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
- CaCO₃: 1 mole* 100 g/mole= 100 g
- NaNO₃: 2 mole* 85 g/mole= 170 g
You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of CaCO₃?
mass of CaCO₃= 74.81 grams
<u><em>74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.</em></u>
The balanced equation is attached in the image below. The coefficients are 2, 2, blank.
