<span>the theoretical yield which is the expected yield and the actual yield obtained are not always the same. therefore percent yield is calculated which shows how much of the percentage of the theoretical yield is actually obtained.
the theoretical yield = 56.0 g
actual yield = 47.0 g
percent yield = actual yield / theoretical yield x 100 %
percent yield = 47.0 / 56.0 x 100% = 83.9 %
percent yield = 83.9 %</span>
1. 2Al(s)+6HCl(aq)⇒2AlCl₃(aq)+3H₂(g)
2. 2AgNO₃ (aq) + Cu (s)⇒Cu(NO₃)₂ (aq) + 2Ag (s)
3. 2C₃H₈O(l) + 9O₂(g) ⇒ 6CO₂(g) + 8H₂O(g)
<h3>Further explanation</h3>
There are several reactions that can occur in a chemical reaction: single replacement, double replacement, synthesis, decomposition or combustion, etc.
1.Al(s)+HCl(aq)⇒AlCl₃(aq)+H₂(g)
type : single replacement
balance :
2Al(s)+6HCl(aq)⇒2AlCl₃(aq)+3H₂(g)
2. AgNO₃ (aq) + Cu (s) ⇒ Cu(NO₃)₂ (aq) + Ag (s)
type : single replacement
balance :
2AgNO₃ (aq) + Cu (s)⇒Cu(NO₃)₂ (aq) + 2Ag (s)
3. C₃H₈O + O₂ ⇒ CO₂ + H₂O
type : combustion of alcohol
balance :
2C₃H₈O(l) + 9O₂(g) ⇒ 6CO₂(g) + 8H₂O(g)
The volume of oxygen at STP required would be 252.0 mL.
<h3>Stoichiometic problem</h3>
The equation for the complete combustion of C2H2 is as below:

The mole ratio of C2H2 to O2 is 2:5.
1 mole of a gas at STP is 22.4 L.
At STP, 100.50 mL of C2H2 will be:
100.50 x 1/22400 = 0.0045 mole
Equivalent mole of O2 according to the balanced equation = 5/2 x 0.0045 = 0.01125 moles
0.01125 moles of O2 at STP = 0.01125 x 22400 = 252.0 mL
Thus, 252.0 mL of O2 gas will be required at STP.
More on stoichiometric problems can be found here: brainly.com/question/14465605
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By crushing the salt, you are performing a physical change because you aren't altering the chemical makeup of the salt, just the physical form. Hope this helps! :)
Answer: Concentration of the chemist's sodium chloride solution is 34.4 mol/L.
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

where,
n= moles of solute
= volume of solution in ml
Given : moles of
= 6.89
volume of solution = 200 ml
Putting in the values we get:

Thus the concentration of the chemist's sodium chloride solution is 34.4 mol/L.