13.5g
Explanation:
Given parameters:
Mass of Na = 10g
Mass of O₂ = 10g
Unknown:
Mass of products formed = ?
Balanced equation = ?
Solution:
The balanced chemical equation is shown below:
4Na + O₂ ⇒ 2Na₂O
In any reaction, the specie in short supply determines the extent of the reaction.
This reaction is not an exclusion. We need to first determine the specie in short supply and use it to estimate the amount of product since we have a 100% yield which signifies that all was used up.
let us convert to moles;
Number of moles of Na = 
= 0.435mole
Number of moles of O₂ = 
= 0.313mole
From the given equation;
4 moles of Na requires 1 mole of O₂;
0.435 moles of Na will require 
= 0.11 moles
But the given amount O₂ is 0.313, this is an excess of 0.313 - 0.11 = 0.203moles
We see that Na is the limiting reagent;
4 moles of Na gives 2 mole of Na₂O
0.435 moles of Na will give 
= 0.22 moles
Mass of Na₂O = number of moles x molar mass = 62 x 0.22 = 13.5g
learn more:
Number of moles brainly.com/question/1841136
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0.008 ÷ 51.3 = 0.0002
Sig Figs
1
0.0002
Decimals
4
0.0002
Scientific Notation
2 × 10-4
E-Notation
2e-4
Words
zero point zero zero zero two
I HOPE I HELP
Physical. When you simply desolve something with water, to don't actually change it. This action can be undone.
An example of a chemical reaction is when the cells within completely change. You you burn wood, you can't go back in time and un-burn it. Does that make sense?
Percentage yield = (actual yield / theoretical yield) x 100%
The balanced equation for the decomposition is,
2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂<span>
O(g)The
stoichiometric ratio between </span>Na₃(CO₃)(HCO₃)·2H₂O(s) and Na₂CO₃(s) is
2 : 3The decomposed mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 1000 kg
= 1000 x 10³ g
Molar mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 226 g mol⁻¹
moles of Na₃(CO₃)(HCO₃)·2H₂O(s) = mass / molar mass
= 1000 x 10³ g / 226 g mol⁻¹
= 4424.78 mol
Hence, moles of Na₂CO₃ formed = 4424.78 mol x
= 6637.17 mol
Molar mass of Na₂CO₃ = 106 g mol⁻¹
Hence, mass of Na₂CO₃ = 6637.17 mol x 106 g mol⁻¹
= 703540.02 g
= 703.540 kg
Hence, the theoretical yield of Na₂CO₃ = 703.540 kg
Actual yield of Na₂CO₃ = 650 kg
Percentage yield = (650 kg / 703.540 kg) x 100%
=
92.34%
