Answer:
11.3 g.
Explanation:
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In this case, since the combustion of butane is:
Thus, since there is a 1:5 mole ratio between butane and water, we obtain the following mass of water:
Therefore, the resulting mass of water is:
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I will present a simple reaction so we can do this conversion:
2H₂ + O₂ → 2H₂O
We will assume we have 32 g of O₂ and we want to find the amount of water, assuming this reaction goes to completion. We must first convert the initial mass to moles, which we do using the molar mass in units of g/mol. The molar mass of O₂ is 32 g/mol.
32 g O₂ ÷ 32 g/mol = 1 mole O₂.
Now that we have moles of oxygen, we use the molar coefficients to find the ratio of water molecules to oxygen molecules. We can see there are 2 moles of water for every 1 mole of oxygen.
1 moles O₂ x (2 mol H₂O/ 1 mol O₂) = 2 moles H₂O
Now that we have the moles of water, we can convert this amount into grams using the molar mass of water, which is 18 g/mol.
2 moles H₂O x 18 g/mol = 36 g H₂O
Now we have successfully converted the mass of one molecule to the mass of another.
First compute the number of grams of manganese metal required to make 21.7 grams of H2.
21.7 g H2 x (1 mole H2/ 2 g H2) x (1 mole Mn/1 mol H2) x (55 grams Mn/1 mol Mn) = 596.75 grams
Now density = mass/volume
7.43 = 596.75/volume
volume = 596.75/7.43 = 80.31 mL
80.31 mL is the amount of manganese needed.
Answer:
In Chemistry, Dobereiner triads are defined as, any of several sets of three chemically same elements, the atomic weight of one, which is nearly equal to the mean of the atomic weights of the other two elements.
