The mass of water produced when 4.86 g of butane(C4H10) react with excess oxygen is calculated as below
calculate the moles of C4H10 used = mass/molar mass
moles = 4.86g/58 g/mol =0.0838 moles
write a balanced equation for reaction
2 C4H10 + 13 O2 = 8 CO2 + 10 H2O
by use of mole ratio between C4H10 to H2O which is 2:10 the moles of
H20= 0.0838 x10/2 = 0.419 moles of H2O
mass = moles x molar mass
=0.419 molx 18 g/mol = 7.542 grams of water is formed
Answer:
Im so confused, and maybe blind bcs I cant read it
Explanation:
Could you explain??
Answer:
<h2>1 N2+ 1 O2 are two N atoms and two O atoms.</h2>
<h3>In NO, there is one N and one O atom.</h3>
For both sides to be the same, there have to be 2 NO molecules, as 2 NO molecules contain 2 N atoms and 2 O atoms, just like starting products of the reaction.
The balanced chemical reaction is:
<span>2 Al (s) + 3 Fe(NO3)2 (aq) = 3 Fe (s) + 2 Al(NO3)2 (aq)
</span>
The ratios or the coefficients between the reactants and the products will be used for further calculations.
305 g Fe(NO3)2 solution (.755) = 230.275 g <span>Fe(NO3)2
</span> 230.275 g Fe(NO3)2 ( 1mol / 179.87 g) ( 3 mol Fe / 3 mol <span>Fe(NO3)2) ( 55.85 g / mol) = 71.500 g Fe</span>
Answer:
C₂H₄O
Explanation:
In a compound that contains Cabon, hydrogen and oxygen, the combustion produce CO₂ from the carbon, and H₂O from the hydrogens. Using the mass of the products we can solve the moles of Carbon and hydrogen. The empirical formula is the simplest whole-number of atoms present in a molecule.
<em>Moles CO₂ = Moles C:</em>
11.8g CO₂ * (1mol / 44g) = 0.268 moles CO₂ = 0.268 moles C * (12g/mol) =
3.216g C
<em>Moles H₂O = 1/2 moles H:</em>
4.83g H₂O * (1mol / 18g) = 0.268 moles H₂O * (2 mol H / 1 mol H₂O) =
0.537 mol H * (1g/mol) = 0.537g H
<em>Mass O to find moles O:</em>
5.90g Sample - 3.216g C - 0.537g H = 2.147g O * (1mol / 16g) = 0.134 moles O
<em>Ratio of atoms -Dividing in 0.134 moles-:</em>
C = 0.268mol C / 0.134 mol O = 2
H = 0.537mol H / 0.134 mol O = 4
O = 0.134mol O / 0.134 mol O = 1
Empirical formula is:
<h3>C₂H₄O</h3>
