How many grams of potassium nitrate (KNO3) are
required to produce 5.00 g of potassium nitrite (KNO2)
according to the equation below?
2KNO3 — 2KNO2 + O2

5.94 g

Explanation:

Looking at the equation, for every 2 moles of KNO₃ added, 2 moles of KNO₂ are produced.  This means that the moles produced will be equal to the moles consumed.

Convert grams of KNO₂ to moles using the molar mass (85.1 g/mol) and get 0.0588 moles of KNO₂.

Since moles of KNO₂ are equal to mole of KNO₃, 0.0588 moles of KNO₃ are required.  Convert this to grams using the molar mass (101.1 g/mol) and get 5.94 g.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

in 1783 cavendish was working with his hydrogen gas and found that when he burned it and condensed the vapor it was water. this was very significant since it disproved the greek theory of elements completely since water was not a simple substance but the product of two gases. it was lavoisier who hearing of the experiment gave cavendish's gas the name hydrogen ("water-producer") and added that hydrogen burned by combining with oxygen and that water was a hydrogen-oxygen combination.

explanation: