Answer:
The volume of the sample is 17.4L
Explanation:
The reaction that occurs requires the same amount of CO and NO. As the moles added of both reactants are the same you don't have any limiting reactant. The only thing we need is the reaction where 4 moles of gases (2mol CO + 2mol NO) produce 3 moles of gases (2mol CO2 + 1mol N2). The moles produced are:
0.1800mol + 0.1800mol reactants =
0.3600mol reactant * (3mol products / 4mol reactants) = 0.2700 moles products.
Using Avogadro's law (States the moles of a gas are directly proportional to its pressure under constant temperature and pressure) we can find the volume of the products:
V1n2 = V2n1
<em>Where V is volume and n moles of 1, initial state and 2, final state of the gas</em>
Replacing:
V1 = 23.2L
n2 = 0.2700 moles
V2 = ??
n1 = 0.3600 moles
23.2L*0.2700mol = V2*0.3600moles
17.4L = V2
<h3>The volume of the sample is 17.4L</h3>
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The answer is the coefficient is "1".
C₅H₁₀, now you see that there is no number with this and when there is no number or digit, it means the coefficient is one.
we use the coefficients to balance the equation of the reaction in such a way that number of the atoms of the elements in the reactants are equal to the number of atoms of different elements in the product, so that both sides are equal and balanced.
Answer:
Explanation:
a )
3NO₂(g) + H₂O(l) — -→ 2HNO₃(aq) + NO(g)
3 x 46 g 18 g 2 x 63 g 30 g
138 g of NO₂ requires 18 g of H₂O
28 g of NO₂ requires ( 18 / 138) x 28
= 3.65 g of H₂O.
b )
18 g of H₂O produces 30 g of NO gas
15.8 g of H₂O produces ( 30/18 ) x 15.8
= 26.33 g of NO gas .
c )
138 g of NO₂ produces 126 g of HNO₃
8.25 g of NO₂ produces (126 / 138 ) x 8.25
= 7.53 g of HNO₃