You carefully weigh out 14.00 g of caco3 powder and add it to 56.70 g of hcl solution. you notice bubbles as a reaction takes place. you then weigh the resulting solution and find that it has a mass of 64.96 g . the relevant equation is
caco3(s)+2hcl(aq)→h2o(l)+co2(g)+cac l2(aq)
assuming no other reactions take place, what mass of co2 was produced in this reaction?

Answer : The mass of produced will be, 6.16 grams.

Explanation : Given,

Mass of = 14 g

Mass of = 56.70 g

Molar mass of = 100 g/mole

Molar mass of = 36.5 g/mole

Molar mass of = 44 g/mole

First we have to calculate the moles of and .

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 1 moles of react with 2 mole of

So, 0.14 moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of .

As, 1 moles of react to give 1 moles of

So, 0.14 moles of react to give 0.14 moles of

Now we have to calculate the mass of .

Therefore, the mass of produced will be, 6.16 grams.

they are both synthetic elements

explanation:

d