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2 years ago

Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)

1 answer:

8 0

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?

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3 years ago

An object of mass 0.5 kg is swung in uniform circular motion. The radius is 2 meters, and the force exerted is 4 N. Calculate th

ZanzabumX [31]

Answer:

8 m/s²

Explanation:

m = 0.5kg

r = 2m

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3 years ago

Construct a three-step synthesis of 3-bromo-3-methyl-2-butanol from 2-methyl-2-butene by dragging the appropriate formulas into

juin [17]

Answer:

See explanation

Explanation:

In order to do this, we need to use 3 reagents to get the final product.

The first one, and logic is the halogenation of the alkene. Doing this, with Br2/CCl4, we'll get an alkane with two bromines, one in carbon 2 and the other in carbon 3.

Then, the next step is to eliminate one bromine of the reactant. The best way to do this, is using sodium ethoxide in ethanol. This is because sodium ethoxide is a relatively strong base, and it will promove the product of elimination in major proportions rather than the sustitution product. If we use NaOH is a really strong base, and it will form another product.

When the sodium ethoxide react, it will form a double bond between carbon 1 and 2 (The carbon where one bromine was with the methyl, changes priority and it's now carbon 3).

The final step, is now use acid medium, such H3O+/H2O or H2SO4/H2O. You can use any of them. This will form an carbocation in carbon 2 (it's a secondary carbocation, so it's more stable that in carbon 1), and then, the water molecule will add to this carbon to form the alcohol.

See the attached picture for the mechanism of this.

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2 years ago

Cars run on gasoline, where octane (C8H18) is the principle component. This combustion reaction is responsible for generating en

Bezzdna [24]

Answer:

10.19 g CO₂

4.69 g H₂O

Explanation:

The combustion reaction of Octane is:

C₈H₁₈ → 8CO₂ + 9H₂O

To calculate the mass of CO₂ and H₂O produced, we need to know the mass of octane combusted.

We calculate the mass of Octane from the given volume and density, using the following conversion factors:

1 gallon = 3.785 L

1 L = 1000 mL

Now we convert 1.24 gallons to mL:

1.24 gallon * $\frac{3.785L}{1gallon} *\frac{1000mL}{1L} =$ 4693.4 mL

We calculate the mass of Octane:

4693.4 mL * 0.703 g/mL = 3.30 g Octane

Now we use the stoichiometric ratios and molecular weights to calculate the mass of CO₂ and H₂O:

CO₂ ⇒ 3.30 g Octane ÷ 114g/mol * $\frac{8molCO_{2}}{1molOctane}$ * 44 g/mol = 10.19 g CO₂

H₂O ⇒ 3.30 g Octane ÷ 114g/mol * $\frac{9molH_{2}O}{1molOctane}$ * 18 g/mol = 4.69 g H₂O

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3 years ago