Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the first heating, your crucible and contents weighs 17.51 g. After the second heating, your crucible and contents weighs 17.50 g. What is the theoretical yield of sodium carbonate

The question is incomplete, the complete question is;

Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the first heating, your crucible and contents weighs 17.51 g. After the second heating, your crucible and contents weighs 17.50 g.

What is the theoretical yield of sodium carbonate?

What is the experimental yield of sodium carbonate?

What is the percent yield for sodium carbonate?

Which errors could cause your percent yield to be falsely high, or even over 100%?

See Explanation

Explanation:

We have to note that water is driven away after the second heating hence we are concerned with the weight of the pure dry product.

Hence;

From the reaction;

2 NaHCO3 → Na2CO3(s) + H2O(l) + CO2(g)

Number of moles of  sodium bicarbonate = 18.56 - 15.98 = 2.58 g/87 g/mol

= 0.0297 moles

2 moles of sodium bicarbonate yields 1 mole of sodium carbonate

0.0297 moles of 0.015 moles  sodium bicarbonate yields 0.0297 * 1/2 = 0.015 moles

Theoretical yield of sodium carbonate = 0.015 moles * 106 g/mol = 1.59 g

Experimental yield of sodium bicarbonate = 17.50 g - 15.98 g = 1.52 g

% yield = experimental yield/Theoretical yield * 100

% yield = 1.52/1.59 * 100

% yield = 96%

The percent yield may exceed 100% if the water and CO2 are not removed from the system by heating the solid product to a constant mass.

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