BRAINLIEST HELP PLEASE. 6. In order to produce CO2, a student put a sample of calcium carbonate in a flask and added hydrochloric acid using the following equation:
CaCO3 + 2HCI- H2O + CO2 + CaCl2
The student finds the mass of CO2 generated in the lab to be 100 g. If the student
began with 200 g of HCl and excess CaCO3, what was the percent yield of CO2?

your answer would be a. 12.5 kg

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

the answer is a for sure mark brainliest i had got this right on usatestprep