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andrew11 [14]
3 years ago
10

How many grams of O2 are needed to react with 8.15 g of C2H2?

Chemistry
1 answer:
xxTIMURxx [149]3 years ago
8 0

Answer:

25.08 grams of O₂ are needed to react with 8.15 g of C₂H₂.

Explanation:

The balanced reaction is:

2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O

By reaction stoichiometry, the following amounts of moles of each compound participate in the reaction:

  • C₂H₂: 2 moles
  • O₂: 5 moles
  • CO₂: 4 moles
  • H₂O: 2 moles

The molar mass of each compound is:

  • C₂H₂: 26 g/mole
  • O₂: 32 g/mole
  • CO₂: 44 g/mole
  • H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • C₂H₂: 2 moles* 26 g/mole= 52 g
  • O₂: 5 moles* 32 g/mole= 160 g
  • CO₂: 4 moles* 44 g/mole= 176 g
  • H₂O: 2 moles* 18 g/mole= 36 g

Then you can apply the following rule of three: if by stoichiometry 52 grams of C₂H₂ react with 160 grams of O₂, 8.15 grams of C₂H₂ react with how much mass of O₂?

mass of O_{2} =\frac{8.15 grams of C_{2} H_{2}*160 grams of O_{2}  }{52 grams of C_{2} H_{2}}

mass of O₂= 25.08 grams

<u><em>25.08 grams of O₂ are needed to react with 8.15 g of C₂H₂.</em></u>

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What mass in grams of MgSO4 is required to make 59.3 mL of 2.68 M<br> solution?
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Answer:

Approximately 19.1\; \rm g.

Explanation:

<h3>Number of moles of formula units of magnesium sulfate required to make the solution</h3>

The unit of concentration in this question is "\rm M". That's equivalent to "\rm mol \cdot L^{-1}" (moles per liter.) In other words:

c(\mathrm{MgSO_4}) = 2.68\; \rm M = 2.68\; \rm mol \cdot L^{-1}.

However, the unit of the volume of this solution is in milliliters. Convert that unit to liters:

\displaystyle V = 59.3\; \rm mL = 59.3 \; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.0593\; \rm L.

Calculate the number of moles of \rm MgSO_4 formula units required to make this solution:

\begin{aligned}n(\rm MgSO_4) &= c \cdot V \\ &= 2.68 \; \rm mol \cdot L^{-1} \times 0.0593\; \rm L \approx 0.159\; \rm mol \end{aligned}.

<h3>Mass of magnesium sulfate in the solution</h3>

Look up the relative atomic mass data of \rm Mg, \rm S, and \rm O on a modern periodic table:

  • \rm Mg: 24.305.
  • \rm S: \rm 32.06.
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Calculate the formula mass of \rm MgSO_4 using these values:

M(\mathrm{MgSO_4}) = 24.305 + 32.06 + 4 \times 15.999 \approx 120.361\; \rm g \cdot mol^{-1}.

Using this formula mass, calculate the mass of that (approximately) 0.159\; \rm mol of \rm MgSO_4 formula units:

\begin{aligned}m(\mathrm{MgSO_4}) &= n \cdot M \\&\approx 0.159 \; \rm mol \times 120.361 \; \rm g \cdot mol^{-1} \approx 19.1\; \rm g\end{aligned}.

Therefore, the mass of \rm MgSO_4 required to make this solution would be approximately 19.1\; \rm g.

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Explanation:

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5 g PRO  × 4 Cal/g =   20

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1. 12 L = 12 dm³

2. 3.18 g

<h3>Further explanation</h3>

Given

1. Reaction

K₂CO₃+2HNO₃⇒ 2KNO₃+H₂O+CO₂

69 g K₂CO₃

2. 0.03 mol/L Na₂CO₃

Required

1. volume of CO₂

2. mass Na₂CO₃

Solution

1. mol K₂CO₃(MW=138 g/mol) :

= 69 : 138

= 0.5

mol ratio of K₂CO₃ : CO₂ = 1 : 1, so mol CO₂ = 0.5

Assume at RTP(25 C, 1 atm) 1 mol gas = 24 L, so volume CO₂ :

= 0.5 x 24 L

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2. M Na₂CO₃ = 0.03 M

Volume = 1 L

mol Na₂CO₃ :

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= 0.03 x 1

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