The amount of Al required will be 15.77 grams
<h3>Stoichiometric problem</h3>
First, the equation of the reaction:
The mole ratio is 2:3.
Mole of 63.0 g of FeO = 63/71.84 = 0.8769 moles
Equivalent moles of Al = 0.8769 x 2/3 = 0.5846 moles
Mass of 0.5846 moles Al = 0.5846 x 26.98 = 15.77 grams
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Answer:
There are 2 nitrogen atoms on the product side.
N2 + 3H2
There are 6 hydrogen atoms on the reactant side.
2NH3
Hope I Helped
but where Is the volume in order for us to determine the concentration. since we have moles in H+ ions
then you can say
concentration = M*1000/V
Answer:
CH₄(g) + 2O₂(g) ---> 1CO₂(g) + 2H₂O(g)
Explanation:
any combustion of a hydrocarbon equation is in form:
CₓHₐ(g) + BO₂(g) ---> YCO₂(g) + ZH₂O(g), where x,a,b,y,z are all whole number positive integers
there will be 1 CO₂ to 2 H₂O, since there is 1 C to 4 H in CH₄; it is not 1:4 since 2 H is needed in H₂O
CH₄(g) + _O₂(g) ---> 1CO₂ + 2H₂O
there is 4 total O on products side, which can make 2O₂
CH₄(g) + 2O₂(g) ---> 1CO₂(g) + 2H₂O(g)
M (C2H6) = n × M = 4.6 × 30 = 13.8 g.
