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11) If 352 mole of butane (CH) is combined with 8.47 moles of oxygen gas, how numerous grams that water will be produced?

117 g


We are offered the amounts of 2 reactants, so this is a limiting reactant problem.

1. We recognize we will require an equation with masses and molar masses, so let’s gather every the info in one place.

M_r: 63.55 32.00 18.02

2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

n/mol: 352 8.47

2. Calculate the quantity of water the each reactant deserve to produce

(a) indigenous C₄H₁₀

Moles that H₂O = 352 mol C₄H₁₀ × (10 mol H₂O/2 mol C₄H₁₀) = 704 mol H₂O

(b) From O₂:

Moles that H₂O = 8.47 mol O₂ × (10 mol H₂O/13 mol O₂) = 6.515 mol H₂O

3. Determine the limiting reactant

The limiting reactant is O₂, because it provides fewer mole of H₂O.

4. Calculation the mass of water

Mass the H₂ = 5.515 mol H₂O × (18.02 g H₂)/(1 mol H₂O) = 117 g H₂O

The reaction produce 117 g H₂O.

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4 0
11 months ago

A 4.00L flask include Ne in ~ 25 C and also 6.00 atm is join by a valve to an 8.00 l flask Ar in ~ 25 C and 2.00 atm.
timama <110>




In this case, because know the volume, temperature and also pressure of the early stage containers, we can compute the moles of every gas prior to the opened of the valve as shown below:


Next, we add them increase to achieve the total moles:


Now, the complete volume:


Finally, the complete pressure is computed by making use of the ideal gas equation:


Best regards!

6 0
8 month ago
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