The balanced chemical reaction is:
<span>2C4H10(g)+13O2(g)->10H2O(g)+8CO2(g)
</span>
<span>Calculate the mass of water produced when 1.77 grams of butane reacts with excessive oxygen?
</span>1.77 g C4H10 (1 mol C4H10/58.14 g C4H10) (10 mol H2O / 2 mol C4H10) ( 1.01 g H2O / 1 mol H2O ) = <span>0.15 g H2O
</span><span>Calculate the mass of butane needed to produce 71.6 of carbon dioxide.
</span>71.6 g CO2 (1 mol CO2/ 44.01 g CO2) ( 2 mol C4H10 / 8 mol CO2 ) (58.14 g C4H10 / 1 mol C4H10 ) = 23.65 g C4H10
<em>V = 151 mL = 151 cm³</em>
<em>d = 0,789 g/mL = 0,789 g/cm³</em>
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d = m/V
m = d×V
m = 0,789×151
<u>m = 119,139g</u>
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Answer:
Moles of NaCl formed is 6.0 moles
Explanation:
We are given the equation;
2 Na(s) + Cl₂(g) → 2 NaCl(s)
- Moles of Na is 6.0 moles
- Moles of Cl₂ is 4.0 moles
From the reaction;
2 moles of sodium reacts with 1 mole of chlorine gas to form 2 moles of NaCl
In this case;
6 moles of Na would require 3 moles of Cl₂, this means that chlorine gas is in excess.
Thus, the rate limiting reagent is sodium.
But, 2 moles of sodium reacts to form 2 moles of NaCl
Therefore;
Moles of NaCl = Moles of Na
= 6.0 moles
Thus, moles of sodium chloride produced is 6.0 moles
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