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ZanzabumX [31]
3 years ago
5

I need help pleasee

Chemistry
1 answer:
Degger [83]3 years ago
5 0

Answer:

We can't see the graph so how are we supposed to help you???

Explanation:

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54.0g Al reacts with 64.0g O2 to form Al2O3 according to the equation.
pentagon [3]

Answer:

136 g Al₂O₃

Explanation:

Assuming you do not need to find the limiting reactant, to find the mass of Al₂O₃, you need to (1) convert grams O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles Al₂O₃ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles Al₂O₃ to grams Al₂O₃ (via molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to match the sig figs of the given value (64.0 g).

Molar Mass (O₂): 32 g/mol

Molar Mass (Al₂O₃): 102 g/mol

4 Al + 3 O₂ -----> 2 Al₂O₃

64.0 g O₂           1 mole           2 moles Al₂O₃            102 g
-----------------  x  --------------  x  ------------------------  x   -------------  =  136 g Al₂O₃
                             32 g               3 moles O₂             1 mole

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Given the following two quantities: 0.50 mol of CH4 and 1.0 mol of HCl,
Vinil7 [7]

Answer:

(a) HCl

(b) HCl

(c) HCl

(d) HCl

Explanation:

<em>Given: </em>0.50 mol of CH₄ and 1.0 mol of HCl

Using stoichiometry we can calculate the answers to parts a, b, c, and d.

<h3>Part (a) </h3>

# of moles × Avogadro's number = # of atoms or molecules

Avogadro's number: 6.02 * 10²³

  • \displaystyle 0.50\ \text{mol CH}_4 \cdot \frac{6.02\cdot 10^2^3 \ \text{atoms CH}_4}{1 \ \text{mol CH}_4} = 3.01 \cdot 10^2^3 \ \text{atoms CH}_4
  • \displaystyle 1.0\ \text{mol HCl} \cdot \frac{6.02\cdot 10^2^3 \ \text{atoms HCl}}{1 \ \text{mol HCl}} = 6.02 \cdot 10^2^3 \ \text{atoms HCl}

HCl has more atoms than CH₄.

<h3>Part (b) </h3>

This is calculated the same way as Part (a); HCl has more molecules than CH₄.

<h3>Part (c) </h3>

Molar mass of CH₄ = 16.04 g/mol

Molar mass of HCl = 36.458 g/mol

  • \displaystyle 0.50\ \text{mol CH}_4 \cdot \frac{16.04 \ \text{g CH}_4}{1 \ \text{mol CH}_4} = 8.02 \ \text{g CH}_4
  • \displaystyle 1.0\ \text{mol HCl} \cdot \frac{36.458 \ \text{g HCl}}{1 \ \text{mol HCl}} = 36.458 \ \text{g HCl}

HCl has a greater mass than CH₄.

<h3>Part (d)</h3>

Assuming STP:

Molar volume of any gas at STP is 22.4 L/mol.

  • \displaystyle 0.50\ \text{mol CH}_4 \cdot \frac{22.4 \ \text{L CH}_4}{1 \ \text{mol CH}_4} = 11.2 \ \text{L CH}_4
  • \displaystyle 1.0\ \text{mol HCl} \cdot \frac{22.4 \ \text{L HCl}}{1 \ \text{mol HCl}} = 22.4 \ \text{L HCl}

HCl has a greater volume than CH₄.

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