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gulaghasi [49]
3 years ago
5

What is the reactant in the following equation?

Chemistry
1 answer:
Degger [83]3 years ago
4 0
To remember what the reactant is, it always comes before the yielding sign (arrow). The elements after the arrow are known as the 'products'.
So, with this information, let's look for our reactants.
Remember, the reactants are the elements BEFORE the yielding sign (arrow).
We have Hydrogen and Oxygen before the arrow.

Your answer is:
Hydrogen + Oxygen

I hope this helps!
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A controlled variable is the variable that is left constant throughout the course of an experiment. The controlled variable here is the gender.

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What mass of chromium would be produced from the reaction of 57.0 g of potassium with 199 g of chromium(II) bromide according to
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Answer:

Mass of Chromium produced = 37.91 grams

Explanation:

2K + CrBr₂  →  2KBr + Cr

2mole     1 mole                1 mole

mass of Potassium = 57.0 grams

molar mass of Potassium = 39.1 g/mol

no of moles of Potassium = 57.0 / 39.1 = 1.458 moles

mass of CrBr₂= 199 grams

molar mass of CrBr₂ = 211.8 gram/mole

no of moles of CrBr₂ = 199 / 211.8 = 0.939 mole

From chemical equation

1 mole of CrBr₂ = 2 moles of K

∴ 0.939 moles of CrBr₂ = ?

   ⇒ 0.939 x 2/1 = 1.878 moles of K

1.878 moles of K is needed, but there is 1.458 moles of K. So, Potassium is completed first during the reaction . Hence, Potassium is limiting reagent. and CrBr₂ is excess reagent .

From chemical equation

2 moles of K = 1 mole of Cr

∴ 1.458 moles of K = ?

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no of moles of Cr formed = 0.729 moles

molar mass of Cr = 52.0 g/mol

mass of one mole of Cr = 52.0 grams

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mass of Chromium produced = 37.91 grams

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3 years ago
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