Calculate the mass of water produced when 42.0 g of propane, c3h8, is burned with 115 g of oxygen. hint: write balanced equation first

            51.75 g of H₂O

Solution:

              The balance chemical equation is as follow,

                                C₃H₈  +  5 O₂    →    3 CO₂  +  4 H₂O

Step 1: Finding Limiting Reagent:

According to equation,

               44.1 g (1 mol) C₃H₈ reacts with  =  160 g (5 mol) of O₂

So,

                        42 g C₃H₈ will react with  =  X g of O₂

Solving for X,

                     X  =  (42 g × 160 g) ÷ 44.1 g

                     X  =  152.38 g of O₂

As we are only provided with 115 g of O₂, Hence it is the limiting reactant and it will control the yield of products.

Step 2: Calculating amount of H₂O:

According to equation,

               160 g (5 mol) O₂ produces  =  72 g (4 mol) of H₂O

So,

                        115 g O₂ will produce  =  X g of H₂O

Solving for X,

                     X  =  (72 g × 115 g) ÷ 160 g

                     X  =  51.75 g of H₂O

answer: a tank of height h is filled with water and sits on the ground. water squirts from a hole at a height y above the ground and has a range r. for two y values, 0 and h, r is zero. determine the value of y for which the range will be a maximum.

explanation:

answer; talc and gypsum;