dissolves in water to give
and
ions according to the following reaction:
-------->
+ 2
So, according to the above reaction, 1 mole of
produce 2 moles of
ion,
So, 0.3 mole will give = 0.3 x 2 = 0.6 moles of
ion
So, Molar concentration =
Note: 1L = 1000mL
The balanced chemical equation would be as follows:
<span>
Mg + O2 → MgO2
</span>We are given the amounts of the reactants. We need to determine first which one is the limiting reactant. We do as follows:
62.0 g Mg (1 mol / <span>24.31 g ) = 2.55 mol Mg
</span>55.5 g O2 ( 1 mol / 32 g ) = 1.74 mol O2 -----> <span>consumed completely and therefore the limiting reactant
2.55 mol - 1.74 mol O2( 1 mol Mg / 1 mol O2 ) = 0.81 mol Mg excess</span>
Hey there!:
moles of AgNO3 = 100 x 0.1 / 1000 = 0.01
moles of NaCl = 100 x 0.200 / 1000 = 0.02
volume of solution = 100 + 100 = 200 mL
mass of solution = 200 x 1 = 200 g
temperature rise = 25.30 - 24.60 = 0.70ºC
Therefore:
Q = m* Cp* ΔT + Cp* ΔT =
200* 4.184 * 0.70 + 15.5 * 0.70 =
Q = 596.61 J
Given the reaction:
Q = 596.61 J
NaCl(aq) + AgNO3(aq)--------> AgCl(s) + NaNO3(aq)
1 mole NaCl ------ 1 mole AgNO3 ------ 1 mole AgCl
0.02 moles NaCl ----- 0.01 moles AgNO3
So:
ΔH = - Q / T
ΔH = - 596.61 * 10⁻³ / 0.01
ΔH = - 59.66 kJ
Hope that helps!
