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2 years ago

In a chemical reaction how does the mass of the reactants compare with the mass of the products?

Chemistry

1 answer:

Juli2301 [7.4K]2 years ago

6 0

Answer:

The mass of the reactants compared with the mass of the products should be the same if the reactants are in stoichiometric amounts.

Explanation:

In this question, they ask about chemical reactions and the comparison of the mass of reactants and products. Firstly, it is necessary to introduce the mass conservation principle.

Mass conservation principle mentions that in a chemical reaction, the total mass of reactants is equal to the total mass of products (if the reaction is fully developed). It means mass is not created or destroyed, only transforms from reactants to products.

For example, the mass of sodium plus the mass of chlorine that reactswith the sodium equals the mass of the product sodium chloride.Because atoms are only rearranged in a chemical reaction, there mustbe the same number of sodium atoms and chlorine atoms in both thereactants and products.

Finally, we can conclude that The mass of the reactants compared with the mass of the products should be the same if the reactants are in stoichiometric amounts.

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2 years ago

A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

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Answer:

pH =1 2.84

Explanation:

First we have to start with the reaction between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

Now for example, we can use a volume of 10 mL of HCl. So, we can calculate the moles using the molarity equation:

$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

$mol=0.282*0.01$

$mol=0.00282$

We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far 0.00282 mol of HCl and 0.00705 mol of KOH. If we check the reaction we have a molar ratio 1:1, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an excess of KOH. This excess can be calculated if we substract the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

Now, if we want to calculate the pH value we will need a concentration, in this case KOH is in excess, so we have to calculate the concentration of KOH. For this, we already have the moles of KOH that remains left, now we need the total volume:

$Total~volume=50mL+10mL=60mL$