Mass C₆H₈O₇ : 0.531484 g
<h3>Further explanation</h3>
Reaction
3NaHCO₃ (aq) + C₆H₈O₇ (aq) → 3 CO₂ (g) + 3 H₂O (l) + Na₃C₆H₅O₇ (aq)
MW NaHCO₃ : 84 g/mol
mass NaHCO₃ : 7.10² mg=0.7 g
mol NaHCO₃ :

mol C₆H₈O₇ :

MW C₆H₈O₇ : 192 g/mol
mass C₆H₈O₇ :

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 heat energy breaks down the bonds between the molecules of the solid so the molecules become looser. Further hearing of the liquid causes the bonds to be broken down and the molecules will move further apart.
To determine the k for the second condition, we use the Arrhenius equation which relates the rates of reaction at different temperatures. We do as follows:
ln k1/k2 = E / R (1/T2 - 1/T1) where E is the activation energy and R universal gas constant.
ln 1.80x10^-2 / k2 = 80000 / 8.314 ( 1/723.15 - 1/593.15)
k2 = 0.3325 L / mol-s
Answer:
5.5 L
Explanation:
First we <u>convert 10 g of propane gas</u> (C₃H₈) to moles, using its <em>molar mass</em>:
- 10 g ÷ 44 g/mol = 0.23 mol
Then we <u>use the PV=nRT formula</u>, where:
- P = 1 atm & T = 293 K (This are normal conditions of T and P)
- R = 0.082 atm·L·mol⁻¹·K⁻¹
1 atm * V = 0.23 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 K