If 3.289 x 10^23 atoms of potassium react with excess water, how many grams of hydrogen gas would be produced?
1 answer:
Answer:
The amount in grams of hydrogen gas produced is 0.551 grams
Explanation:
The parameters given are;
Number of atoms of potassium, aₙ = 3.289 × 10²³ atoms
Chemical equation for the reaction is given as follows;
2K + 2H₂O KOH + H₂
Avogadro's number, , regarding the number of molecules or atom per mole is given s follows;
= 6.02 × 10²³ atoms/mole
Therefore;
The number of moles of potassium present = 3.289 × 10²³/(6.02 × 10²³) = 0.546 moles
2 moles of potassium produces one mole of hydrogen gas, therefore;
1 moles of potassium produces 1/2 mole of hydrogen gas, and 0.546 moles of potassium will produce 0.546/2 moles of hydrogen which is 0.273 moles of hydrogen gas
The molar mass of hydrogen gas = 2.016 grams
Therefore, 0.273 moles will have a mass of 0.273×2.016 = 0.551 grams.
The amount in grams of hydrogen gas produced = 0.551 grams.
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Explanation:
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In this case, according to the given information, it turns out possible for us to solve for the molarity of the KOH solution by knowing this base react in a 1:1 mole ratio with nitric acid, HNO3; thus, we can write the following equation, as their moles are the same at the endpoint:
Which in terms of molarities and volumes is:
Thus, we solve for the molarity of the base (KOH) to obtain:
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