Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)
Starting with with 200.0 grams of Pb(NO3)2 and 120.0 grams of NaI:
A. What is the limiting reagent?
B. How many grams of PbI2 is theoretically formed?
C. How many grams of the excess reactant remains?
D. If 48 grams of NaNO3 actually formed in the reaction, what is the percent yield of this reaction?
The number of mole sulphuric acid in each mL of solution is 0.0183 mol/mL.
<h3>What is concentration?</h3>
- Concentration in chemistry is calculated by dividing a constituent's abundance by the mixture's total volume.
- Mass concentration, molar concentration, number concentration, and volume concentration are four different categories of mathematical description.
- Any type of chemical mixture can be referred to by the term "concentration," however solutes and solvents in solutions are most usually mentioned.
- There are different types of molar (quantity) concentration, including normal concentration and osmotic concentration.
<h3>How is concentration determined?</h3>
- Subtract the solute's mass from the total volume of the solution. Using m as the solute's mass and V as the total volume of the solution, write out the equation C = m/V.
- To get the concentration of your solution, divide the mass and volume figures you discovered and plug them in.
Learn more about concentration here:
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Answer:
37.1°C.
Explanation:
- Firstly, we need to calculate the amount of heat (Q) released through this reaction:
<em>∵ ΔHsoln = Q/n</em>
no. of moles (n) of NaOH = mass/molar mass = (2.5 g)/(40 g/mol) = 0.0625 mol.
<em>The negative sign of ΔHsoln indicates that the reaction is exothermic.</em>
∴ Q = (n)(ΔHsoln) = (0.0625 mol)(44.51 kJ/mol) = 2.78 kJ.
Q = m.c.ΔT,
where, Q is the amount of heat released to water (Q = 2781.87 J).
m is the mass of water (m = 55.0 g, suppose density of water = 1.0 g/mL).
c is the specific heat capacity of water (c = 4.18 J/g.°C).
ΔT is the difference in T (ΔT = final temperature - initial temperature = final temperature - 25°C).
∴ (2781.87 J) = (55.0 g)(4.18 J/g.°C)(final temperature - 25°C)
∴ (final temperature - 25°C) = (2781.87 J)/(55.0 g)(4.18 J/g.°C) = 12.1.
<em>∴ final temperature = 25°C + 12.1 = 37.1°C.</em>
I think the answer is a pure substance
