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2 years ago

What volume in milliliters of 0.100 M HCIO3 is required to neutralize 40.0 mL of 0.165 M KOH?

Chemistry

1 answer:

Veronika [31]2 years ago

7 0

Answer:

66.0 mL HClO3

Explanation:

M1V1 = M2V2

M1 = 0.100 M HClO3

V1 = ?

M2 = 0.165 M KOH

V2 = 40.0 mL KOH

Solve for V1 --> V1 = M2V2/M1

V1 = (0.165 M)(40.0 mL) / (0.100 M) = 66.0 mL HClO3

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3 years ago

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2 years ago

1.5mol C3H8 from C3H8+5O2-->3CO2+4H2O .how many grams of carbon dioxide are produced

koban [17]

Answer:

$\large \boxed{\text{200 g CO}_{{2}}}$

Explanation:

We will need a balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

Mᵣ: 44.01

C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

n/mol: 1.5

1. Calculate the moles of CO₂

The molar ratio is 3 mol CO₂:1 mol C₃H₈

$\rm \text{Moles of CO}_{2} = \text{1.5 mol C$_{3}$H}_{8} \times \dfrac{\text{3 mol CO}_{2}}{\text{1 mol C$_{3}$H}_{8}} =\text{4.5 mol CO}_{2}$

2. Calculate the mass of CO₂.

$\text{Mass of CO}_{2} = \text{4.5 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO$_{2}$}} = \textbf{200 g CO}_{\mathbf{2}}\\\text{The reaction will form $\large \boxed{\textbf{200 g CO}_{\mathbf{2}}}$}$

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2 years ago