Tin(IV) sulfide, SnS 2 , a yellow pigment, can be produced using the following reaction. SnBr 4 ( aq ) + 2 Na 2 S ( aq ) ⟶ 4 NaBr ( aq ) + SnS 2 ( s ) Suppose a student adds 47.7 mL of a 0.474 M solution of SnBr 4 to 43.4 mL of a 0.179 M solution of Na 2 S . Identify the limiting reactant.

The limiting reactant is Na2S.

We have the equation of the reaction as follows;

SnBr4 ( aq ) + 2Na2S ( aq ) ⟶ 4NaBr ( aq ) + SnS2 ( s )

Number of moles of SnBr4 = concentration × volume

= 0.474 M ×  47.7/1000 L = 0.023 moles

Number of moles of Na2S = 0.179 M  × 43.4/1000 L = 0.0078 moles

From the reaction equation;

1 mole of SnBr4 reacts with 2 moles of Na2S

0.023 moles of SnBr4 reacts with 0.023 moles × 2 moles/1 mole

= 0.046 moles

We can see that the amount of Na2S required for the reaction is far greater than what is available hence there is not enough Na2S present. Na2S is the limiting reactant.

Learn more: link

is the limiting reagent

Explanation:

To calculate the number of moles for given molarity, we use the equation:

    .....(1)

1. Molarity of solution = 0.474 M

Volume of solution = 47.7 mL = 0.0477 L

Putting values in equation 1, we get:

2. Molarity of solution = 0.179 M

Volume of solution = 43.4 mL = 0.0434 L

Putting values in equation 1, we get:

According to stoichiometry :

2 moles of require 1 mole of

Thus 0.00777 moles of will require=  of

Thus is the limiting reagent as it limits the formation of product and is the excess reagent.

the answer is a. co2, this would be colder than the : )

explanation:

answer: semi-infinite solid initially at a uniform temperature of 400 ◦c is cooled at its surface by convection. the cooling fluid at a constant temperature of 200 ◦c has a convective coefficient (ha) of 250 w/m2.k. the physical properties of the solid: k = 20 w/m.k and α = 4 * 10-5  m2/s. using δx = 0.04 m and δt = 10 s. write the finite difference equations describing the transient heat transfer of the first five δx and the temperature profile after 30 s.