The products are Potassium bromide(KBr), Water(H2O) and Carbon(CO2).
KHCO3 + HBr ----> KBr + H2O + CO2
Bromine reacts with water to form a mixture of Hydrobromic Acid, HBr, and Hypobromous Acid, HBrO.
The moles of KHCO3 and KCl produced should be the same because they are stoichiometrically related in the chemical reaction that produces them. For every mole of KHCO3 that reacts, it produces one mole of KCl. This means that the number of moles of KHCO3 consumed is equal to the number of moles of KCl produced in the reaction.
When potassium bromide (KBr) reacts with hydrochloric acid (HCl), a double displacement reaction occurs producing potassium chloride (KCl) and hydrogen bromide (HBr). The reaction can be represented as follows: KBr + HCl → KCl + HBr.
The mass of KCl recovered is less than the starting mass of KHCO3 because during the chemical reaction between KHCO3 and HCl to form KCl, CO2 gas is evolved. Some of the starting mass of KHCO3 is lost as gas during the reaction, leading to a lower mass of the end product (KCl) compared to the original mass of KHCO3.
HBr + NaOH ------> NaBr + H2O This is an acid-base reaction. The compounds will disassociate into ions in solution. The hydrogen from the HBr will go to the OH- and form water. The NaBr is a salt.
Potassium bicarbonate (KHCO3) and hydrobromic acid (HBr) will react to form potassium bromide (KBr), water (H2O), and carbon dioxide (CO2). Here's the balanced equation:KHCO3 + HBr --> KBr + H2O + CO2(A closer look actually tells you that the other product besides potassium bromide is carbonic acid, H2CO3; but under normal conditions, that quickly decomposes into water and carbon dioxide, so the above equation actually reflects two different reactions: one double replacement and one decomposition.)
HBR doesn't react with Propane, but it does with Propene. The product is either 1-bromo propane(minor product) or 2-bromo propane(major product). To determine which product will be the major product, use the Markovnikov's rule.
Bromine reacts with water to form a mixture of Hydrobromic Acid, HBr, and Hypobromous Acid, HBrO.
The moles of KHCO3 and KCl produced should be the same because they are stoichiometrically related in the chemical reaction that produces them. For every mole of KHCO3 that reacts, it produces one mole of KCl. This means that the number of moles of KHCO3 consumed is equal to the number of moles of KCl produced in the reaction.
When 1-propene reacts with hydrogen bromide (HBr), it undergoes an electrophilic addition reaction, yielding 2-bromopropane as the major product. This reaction follows Markovnikov's rule, where the bromine atom attaches to the more substituted carbon atom of the double bond. Additionally, a minor product, 1-bromopropane, may also be formed.
In the experiment, excess HCl was added to ensure that all KHCO3 was reacted. Excess reactants ensure that all of the limiting reactant is completely consumed in the reaction, leaving none unreacted. When all KHCO3 reacts, the reaction reaches completion.
When potassium bromide (KBr) reacts with hydrochloric acid (HCl), a double displacement reaction occurs producing potassium chloride (KCl) and hydrogen bromide (HBr). The reaction can be represented as follows: KBr + HCl → KCl + HBr.
The mass of KCl recovered is less than the starting mass of KHCO3 because during the chemical reaction between KHCO3 and HCl to form KCl, CO2 gas is evolved. Some of the starting mass of KHCO3 is lost as gas during the reaction, leading to a lower mass of the end product (KCl) compared to the original mass of KHCO3.
Bromomethane CHBr3 and Hydrogen Bromide HBr
When bromine reacts with water, it forms hydrobromic acid (HBr) and hypobromous acid (HOBr). The overall reaction can be represented as: Br2 + H2O → HBr + HOBr. This reaction is reversible and depends on the pH and conditions of the solution.
HBr + NaOH ------> NaBr + H2O This is an acid-base reaction. The compounds will disassociate into ions in solution. The hydrogen from the HBr will go to the OH- and form water. The NaBr is a salt.
The reaction is:Cd + 2 HBr = CdBr2 + H2