For each of the reactions, calculate the mass (in grams) of the product formed when 15.16 g of the underlined reactant completely reacts. assume that there is more than enough of the other reactant.
a.- 2k(s) + cl2(g) ? 2kcl(s)
b.- 2k(s) + br2(l) ? 2kbr(s)
c.- 4cr(s) + 3o2(g) ? 2cr2o3(s)
d. - 2sr(s) +o2(g) ? 2sro(s)

A. 28.98g

B. 46.25g

C. 22.192g

D. 20.64g

Explanation:

To calculate the masses, all we need to know is the theoretical number of moles and the actual number of moles of the compound given and the one we’re asked to find. We proceed as follows:

A. Here, 2 moles of potassium metal gave 2 moles of potassium chloride. This means that their number of moles are equal.

Now we need to calculate the number of moles of potassium reacted. This is equal to the mass of potassium reacted divided by the atomic mass of potassium.

The atomic mass of potassium is 39.

Thus the number of moles reacted equals 15.16/39 = 0.389 moles

Since we know that the number of moles are equal, it means the number of moles of potassium chloride yielded is also 0.389 moles.

To calculate the mass of potassium chloride yielded, we simply multiply the number of moles of potassium chloride by the molar mass of the potassium chloride. The molar mass of the potassium chloride is 39 + 35.5 = 74.5g/mol.

The mass thus produced is = 0.389 * 74.5 = 28.98g

B. Here, 2 moles of potassium metal yielded 2 moles of potassium bromide. Now , we will need to calculate the actual number of moles of potassium bromide yielded.

We first need to calculate the number of moles of potassium yielded and it can be calculated by dividing the mass of the potassium metal by the atomic mass. This equals 15.16/39= 0.389 moles

We have established from above that 0.389 moles of potassium bromide was also produced. To get the mass of potassium bromide produced, we simply multiply the number of moles by the molar mass of potassium bromide. The molar mass of potassium bromide is 39 + 79.9 = 118.9g/mol

The mass thus produced is number of moles multiplied by the molar mass= 118.9 * 0.389 = 46.25g

C. Here 4 moles of chromium yielded 2 moles of chromium iii oxide theoretically. This means we will have to divide the actual number of moles of chromium produced by 2.

To get the actual number of moles of chromium produced, we simply divide the mass by the atomic mass. The atomic mass of chromium is 52.

The number of moles thus produced is 15.16/52 = 0.292 moles

Now we need to half this number to get the actual number of moles of chromium iii oxide produced as the mole ratio is 2 to 1

Hence, the actual number of moles of chromium produced is 0.292/2 = 0.146 moles

To get the mass of chromium iii oxide produced, we simply multiply the number of moles by the molar mass. The molar mass of chromium iii oxide is = 152g/mol

The mass thus produced is 152 * 0.146 = 22.192g

D. Here 2 moles of strontium gave 2 moles of strontium ii oxide. This means that their number of moles are the same.

Now we need to know the actual number of moles of strontium produced. We get this by dividing the mass of strontium by the molar mass. The mass of strontium is 88g/mol

Number of moles = 15.16/88 = 0.172

Now we have established that their number of moles are the asme. This means 0.172 moles of strontium ii oxide was produced.

The mass thus produced is the number of moles of strontium ii oxide multiplied by its molar mass The molar mass is 88 + 32= 120g/mol

The mass is now-: 120 * 0.172 =. 20.64g

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

58750.08068 is the answer i think