Explain how percent yield is calculated

If 25 mL of a HCl solution of unknown concentration was neutralized with 10 mL of a 0.30 M NaOH solution, what was the original

QveST [7]

Answer:

0.12 M

Explanation:

Step 1: Write the balanced equation

NaOH + HCl ⇒ NaCl + H₂O

Step 2: Calculate the reacting moles of NaOH

10 mL of a 0.30 M NaOH solution react.

$0.010L \times \frac{0.30mol}{L} = 3.0 \times 10^{-3} mol$

Step 3: Calculate the reacting moles of HCl

The molar ratio of NaOH to HCl is 1:1. The reacting moles of HCl are 1/1 × 3.0 × 10⁻³ mol = 3.0 × 10⁻³ mol.

Step 4: Calculate the concentration of HCl

3.0 × 10⁻³ mol of HCl are in 25 mL of solution.

$M = \frac{3.0 \times 10^{-3} mol}{0.025L} = 0.12 M$

5 0

3 years ago

A 3.8-mol sample of KClO3 was decomposed according to the equation. How many moles of O2 are formed assuming 100% yield?

kari74 [83]

Answer:

5.7 moles of O2

Explanation:

We'll begin by writing the balanced decomposition equation for the reaction. This is illustrated below:

2KClO3 —> 2KCl + 3O2

From the balanced equation above,

2 moles of KClO3 decomposed to produce 3 moles of O2.

Next, we shall determine the number of mole of O2 produced by the reaction of 3.8 moles of KClO3.

Since 100% yield of O2 is obtained, it means that both the actual yield and theoretical yield of O2 are the same. Thus, we can obtain the number of mole of O2 produced as follow:

From the balanced equation above,

2 moles of KClO3 decomposed to produce 3 moles of O2.

Therefore, 3.8 moles of KClO3 will decompose to produce = (3.8 × 3)/2 = 5.7 moles of O2.

Thus, 5.7 moles of O2 were obtained from the reaction.

3 0

3 years ago

When 7.80 mL of 0.500 M AgNO3 is added to 6.25 mL of 0.300 M NH4Cl, how many grams of AgCl are formed?

irina1246 [14]

Answer:

The answer to your question is 0.269 grams of AgCl

Explanation:

Data

[AgNO₃] = 0.50 M

Vol AgNO₃ = 7.80 ml

[NH₄Cl] = 0.30 M

Vol NH₄Cl = 6.25 ml

mass of AgCL

Balanced reaction

AgNO₃(aq) + NH₄Cl(aq) ⇒ AgCl (s) + NH₄NO₃ (aq)

Process

1.- Calculate the moles of AgNO₃

Molarity = moles / volume

moles = Molarity x volume

moles = 0.50 x 0.0078

moles = 0.0039

2.- Calculate the moles of NH₄Cl

moles = 0.30 x 0.0063

moles = 0.00188

3.- Calculate the limiting reactant

The proportion of AgNO₃(aq) to NH₄Cl(aq) is 1 :1, then, we conclude that the limiting reactant is NH₄Cl(aq), because there are less amount of this reactant in the experiment.

4.- Calculate the moles of AgCl

1 mol of NH₄Cl ---------------- 1 mol of AgCl

0.00188 mol of NH₄Cl ------------- x

x = (0.00188 x 1) /1

x = 0.00188 moles of AgCl

5.- Calculate the grams of AgCl

molecular mass of AgCl = 108 + 35.5 = 143.5 g

143.5 grams of AgCl -------------- 1 mol

x -------------------------------------------0.00188 moles of AgCl

x = (0.00188 x 143.5) / 1

x = 0.269 grams of AgCl

8 0

3 years ago

An organic compound composed of carbon and hydrogen connected only by single bonds is an ________. alkane alkene alkyne aromatic

My name is Ann [436]

Alkane, the others have double covalent bonds

4 0

3 years ago

Plz help me with this:

In-s [12.5K]

Answer:

1) H 1s^1

2) [Ar] 3d⁸ 4s²

3) [He] 2s2 2p3

4) [Kr] 4d10 5s2 5p5

5) [Ar] 4s² and [Ar] 4s²

6) [He] 2s2 2p2

7) [He] 2s² 2p⁴

8) [Ar] 3d7 4s2

9) [Kr] 4d¹⁰ 5s¹

10) [Ne] 3s² 3p⁶

Explanation:

7 0

3 years ago