<span> Mg(OH)2(s) + 2HCl(aq) yield MgCl2(aq) + 2H2O(l)
grams HCl required = (50.6 grams Mg(OH)2) * (1 mol Mg(OH)2 / 58.3197 grams Mg(OH)2) * (2 mol HCl / 1 mol Mg(OH)2) * (36.453 grams HCl / 1 mol HCl) = 63.26 grams HCl required
Since there are only 45.0 grams HCl, then HCl is the limiting reactant.
theoretical yield MgCl2 = (45.0 grams HCl) * (1 mol HCl / 36.453 grams HCl) * (1 mol MgCl2 / 2 mol HCl) * (95.211 grams MgCl2 / 1 mol MgCl2) = 58.6 grams MgCl2 </span>
The moles of oxygen that are produced when 26.5 mol of Al2O3 decomposes is 39.8 mol
<u>calculation</u>
<u> </u> 2Al2O3 + 4Al +3 O2
- use the mole ratio of Al2O3 to O2 to determine the moles of O2.
- that is from the equation above the mole ratio of Al2O3 : O2 is 2:3
- the moles of O2 is therefore=n 26.5 mol x3/2= 39.8 moles
When the concentration is expressed in percentage, you simply have to divide the amount of substance to the total amount of the mixture, then multiply it by 100. In this case, when you want to find the percentage of P in the sample, the solution is as follows:
%P = 7.5 g/100 g * 100 =<em> 7.5%</em>
The answer is D; Mercury-194
All of the others are not when I looked them up
Answer:
g NaCl = 424.623 g
Explanation:
<em>C</em> NaCl = 3.140 m = 3.140 mol NaCl / Kg solvent
∴ solvent: H2O
∴ mass H2O = 2.314 Kg
mol NaCl:
⇒ mol NaCl = (3.140 mol NaCl/Kg H2O)×(2.314 Kg H2O) = 7.266 mol NaCl
∴ mm NaCl = 58.44 g/mol
⇒ g NaCl = (7.266 mol NaCl)×(58.44 g/mol) = 424.623 g NaCl
