Question

For each of the following unbalanced equations, calculate how many grams of each product would be produced by complete reaction of 12.5 g of the reaction indicated in boldface, indicate clearly the mole ration used for the conversion.a. TiBr₄(g) + H₂(g) $\rightarrow$ Ti(s) + HBr(g)
b. SiH₄(g) + NH₃(g) $\rightarrow$ Si3N₄(s) + H₂(g)

Answer 1

Answer:

The answer to your question is

Explanation:

Data

12.5 g of reactant

Balanced Reaction 1

                 TiBr₄ + 2H₂  ⇒  Ti  +  4HBr

Molar mass of TiBr₄ = 48 + (4 x 80) = 368 g

Atomic mass of Ti = 48 g

Molar mass of HBr = 1 + 80 = 81

                 368 g of TiBr₄ ---------------- 48 g of Ti

                    12.5 g of TiBr₄ -------------- x

                    x = (12.5 x 48) / 368

                   x = 1.63 g of Ti

                  368 g of TiBr₄ ----------------4(81) g of HBr

                    12.5 g of TiBr₄ -------------  x

                    x = (12.5 x 324) / 368

                    x = 11 g of HBr

Balanced reaction 2

                  3SiH₄  +  4NH₃  ⇒   Si₃N₄  +  12H₂

Molar mass of SiH₄ = 28 + 4 = 32

Molar mass of Si₃N₄ = 28 x 3 + 14 x 4 = 84 + 56 = 140 g

Atomic mass of H₂ = 2 g

                  3(32) g of SiH₄ --------------- 140 g of Si₃N₄

                    12.5 g of SiH₄ --------------  x

                    x = 18.2 g of Si₃N₄

                  3(32) g of SiH₄ --------------- 24 g of H₂

                  12.5 g of SiH₄   --------------  x

                   x = 3.125 g of H₂