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ivanzaharov [21]
3 years ago
5

Please help me it’s just that I can’t remember how to do this it’s about the Mole and Stoichiomerty

Chemistry
1 answer:
joja [24]3 years ago
8 0

Answer:

6.52moles of H2O

Explanation:

First, let us calculate the number of moles of O2 containing 1.96x10^24 molecules. This is illustrated below:

1mole of O2 contains 6.02x10^23 molecules.

Therefore Xmol of O2 contain 1.96x10^24 molecules i.e

Xmol of O2 = 1.96x10^24/6.02x10^23 = 3.26moles

Now let us generate a balanced equation for the formation of water. This is illustrated below:

2H2 + O2 —> 2H2O

From the equation,

1mole of O2 produced 2moles of H2O.

Therefore, 3.26moles of O2 will produce = 3.26 x 2 = 6.52moles of H2O

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<em>Note: the complete second part of the question is given below:</em>

<em>2. Suppose 0.470 kg of nonane are burned in air at a pressure of exactly 1 atm and a temperature of 17.0 °C. Calculate the volume of carbon dioxide gas that is produced. Round your answer to 3 significant digits.</em>

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C₉H₂₀ (l) + 14O₂ (g) ----> 9CO₂ (g) + 10H₂0 (g)

Part 2: volume of carbon dioxide produced

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At s.t.p., I mole of  C₉H₂₀ reacts with 14 moles of O₂ to produce 9 moles of CO₂

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470 g of C₉H₂₀ will produce 470 * (313.6/128) L of CO₂ = 1151.50 L of CO₂

Volume of CO₂ gas produced at 1 atm and 17 °C;

Using P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁T₂/P₂T₁

where P₁ = 1 atm, V₁ = 1151.50 L, T₁ = 273 K, P₂ = 1 atm, T₂ = 17 + 273 = 290 K

Substituting the values; V₂ = (1 * 1151.5 * 290)/(1 * 273)

Therefore volume of CO₂ produced, V₂ = 1223.21 L of CO₂

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