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olga2289 [7]
3 years ago
6

Trisilane (Si3H8) is a liquid with a density of 0.739 g cm-3. It reacts with oxygen to give silicon dioxide (SiO2) and water. Ca

lculate the mass of water that would form if 20.0 cm3 of trisilane reacted completely with excess oxygen.
Chemistry
1 answer:
MA_775_DIABLO [31]3 years ago
3 0

<u>Answer:</u> The mass of water that would form is 11.52 grams

<u>Explanation:</u>

To calculate the mass of trisilane, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of trisilane = 0.739g/cm^3

Volume of trisilane = 20.0cm^3

Putting values in above equation, we get:

0.739g/cm^3=\frac{\text{Mass of trisilane}}{20.0cm^3}\\\\\text{Mass of trisilane}=(0.739g/cm^3\times 20.0cm^3)=14.78g

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of trisilane = 14.78 g

Molar mass of trisilane = 92.3 g/mol

Putting values in equation 1, we get:

\text{Moles of trisilane}=\frac{14.78g}{92.3g/mol}=0.160mol

The chemical equation for the reaction of trisilane and oxygen gas follows:

Si_3H_8+5O_2\rightarrow 3SiO_2+4H_2O

By Stoichiometry of the reaction:

1 mole of trisilane produces 4 moles of water

So, 0.160 moles of trisilane will produce = \frac{4}{1}\times 0.160=0.64mol of water

Now, calculating the mass of water by using equation 1, we get:

Moles of water = 0.64 moles

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

0.64mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.64mol\times 18g/mol)=11.52g

Hence, the mass of water that would form is 11.52 grams

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gulaghasi [49]
<h3>Answer:</h3>

3.38 × 10²⁴ molecules CO₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 5.61 moles CO₂

[Solve] molecules CO₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                      \displaystyle 5.61 \ mooles \ CO_2(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                          \displaystyle 3.37834 \cdot 10^{24} \ molecules \ CO_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

3.37834 × 10²⁴ molecules CO₂ ≈ 3.38 × 10²⁴ molecules CO₂

6 0
3 years ago
When maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas write the balanved equati
babunello [35]

Answer: When maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}.

Explanation:

The word equation is given as maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas.

Now, in terms of chemical formulae this reaction equation will be as follows.

MnO_{2} + HCl \rightarrow H_{2}O + MnCl_{2} + Cl_{2}

Here, number of atoms on reactant side are as follows.

  • Mn = 1
  • O = 2
  • H = 1
  • Cl = 1

Number of atoms on product side are as follows.

  • Mn = 1
  • O = 1
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  • Cl = 4

To balance this equation, multiply HCl by 4 on reactant side and multiply H_{2}O by 2 on product side. Therefore, the equation can be rewritten as follows.

MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}

Hence, number of atoms on reactant side are as follows.

  • Mn = 1
  • O = 2
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Number of atoms on product side are as follows.

  • Mn = 1
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  • H = 4
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Since, this equation contains same number of atoms on both reactant and product side. Therefore, this equation is now balanced equation.

Thus, we can conclude that when maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}.

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