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Burka [1]
4 years ago
12

For the reaction, calculate how many grams of the product form when 14.4 g of Br2 completely reacts. Assume that there is more t

han enough of the other reactant. 2K(s)+Br2(l)→2KBr(s)
Chemistry
1 answer:
lora16 [44]4 years ago
7 0

Answer:

The answer to your question is 21.45 g of KBr

Explanation:

Chemical reaction

                               2K + Br₂   ⇒   2KBr

                                       14.4            ?

Process

1.- Calculate the molecular mass of bromine and potassium bromide

Bromine = 2 x 79.9 = 159.8g

Potassium bromide = 2(79.9 + 39.1) = 238 g

2.- Solve it using proportions

              159.8 g of Bromine ------------ 238 g of potassium bromide                    

                14.4 g of Bromine  ------------  x

                        x = (14.4 x 238) / 159.8

                        x = 3427.2 / 159.8

                        x = 21.45g of KBr

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What mass of CaSO3 must have been present initially to produce 14.5 L of SO2 gas at a temperature of 12.5°C and a pressure of 1.
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When the reaction equation is:

CaSO3(s) → CaO(s) + SO2(g)

we can see that the molar ratio between CaSO3 & SO2 is 1:1 so, we need to find first the moles SO2.

to get the moles of SO2 we are going to use the ideal gas equation:

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n is the moles' number (which we need to calculate)

R ideal gas constant = 0.0821

and T is the temperature in Kelvin = 12.5 + 273 = 285.5 K

so, by substitution:

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so we can easily get the mass of CaSO3:

when mass = moles * molar mass

and we know that the molar mass of CaSO3= 40 + 32 + 16 * 3 = 120 g/mol


∴ mass = 0.68 moles* 120 g/mol = 81.6 g
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