Calculating Zn Required for 2.00g of Au in Chemical Reaction

[gokugreene]

Hey guys could you tell me if I got the right answer; my teacher hasn't went over this material and it is due tomorrow.

Given the following reaction, how many grams of Zn are required to obtain 2.00g of Au?
$$4CN^{-1}~+~3Zn~+~2Au(CN)_4^{-1}~~-->~3Zn(CN)_4^{-2}~+~2Au$$

(2.00g of Au)(1 mole Au/179.0g of Au)(3 mole Zn/2 mole Au)(65.39g Zn/1 mole Zn) = 1.10g of Zn

Is this the proper way and is my answer correct? Thanks

 

[stunner5000pt]

taht is the correct way of doing it, just be careful of ratios always

 

[dextercioby]

And one more thing.Use the same # of significant digits when working with molar masses.

Daniel.