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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Answer:
209.98 g of NaOH
Explanation:
We are given;
- Volume of HCl as 3 L
- Molarity of HCl as 1.75 M
We are required to calculate the mass of NaOH required to completely neutralize the acid given.
First, we write a balanced equation for the reaction between NaOH and HCl
That is;
NaOH + HCl → NaCl + H₂O
Second, we determine the number of moles of HCl
Number of moles = Molarity × Volume
= 1.75 M × 3 L
= 5.25 moles
Third, we use the mole ratio to determine the moles of NaOH
From the reaction,
1 mole of NaOH reacts with 1 mole of HCl
Therefore;
Moles of NaOH = Moles of HCl
= 5.25 moles
Fourth, we determine the mass of NaOH
Molar mass of NaOH = 39.997 g/mol
Mass of NaOH = 5.25 moles × 39.997 g/mol
= 209.98 g
Thus, 209.98 g of NaOH will completely neutralize 3L of 1.74 M HCl
<span> Ag(NH3)2Cl + 3HNO3 = AgNO3 +2NH4NO3 + HCl </span>
<span>or
Ag(NH3)2Cl + HNO3 = Ag(NH3)2NO3 + HCl this the complete balanced equation
now remove spectator ions to get net ionic equation
so
</span>
<span>
2H+ + 2NO3- + [Ag(NH3)2]+ Cl- -> AgCl + 2NH4+ + 2NO3- 2NO3- 2H+ [Ag(NH3)2]+ + Cl- -> AgCl + 2NH4+
</span>hope it helps
