What is the percent composition (by mass) of MgBr2?

13.2% Mg, 86.8% Br

Explanation:

Mg has a molar mass of 24.31g/mol.

Bromine has a molar mass of 79.90 g/mol.

There is one mole of Mg and two moles of Br in the compound:

79.90x2 + 24.31 = 184.11

So the total molar mass of MgBr2 is 184.11.

A percent tells us a part of a whole, so we will divide each part of the mass into the whole mass.

24.31g Mg / 184.11g MgBr2 = .132

159.8g Br / 184.11g MgBr2 = .868

Multiply these decimals by 100 to make them into a percent.

.132 x 100 = 13.2% Mg

.868 x 100 = 86.8% Br

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

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hope it ,

bioteacher101