Chemistry, PLEASE HELP ASAP! Acetylene gas, also known as ethyne, is used in welding and for many other industrial applications

Question

Acetylene gas, also known as ethyne, is used in welding and for many other industrial applications

a) if 13.7g of ethyne, C2H2 reacts with 18.5g O2 producing carbon dioxide and water vapor, what is the limiting reagent?

b) What is the amount in moles of CO2 that can be produced?

c) What mass is this?

d) What mass of excess reagent remains?

Answer 1

2C2H2+5O2→4CO2+2H2O

a)

have 13.7 g C2H2

have 18.5 g O2

how many grams of O2 are needed?

molar mass

13.7gC2H2/34.04g=0.402mol C2H2

use coefficients

0.402 mol C2H2 (5molO2/2molC2H2)=1.0062 mol O2

use molar mass

1.0062 mol O2(32.00gO2/1molO2)=32.197gO2 needed

have 18.5 g O2

need 32.197 g O2

not enough O2

O2 is limiting

b)

Use limiting reactant amount available (18.5 g O2)

use molar mass

18.5gO2(1molO2/32gO2)=0.578125 mol O2

use coefficients

0.578125 mol O2(4molCO2/5molO2)=0.4625mol CO2

c)

use molar mass

0.4625mol CO2(44.01gCO2/1molCO2)=2.04 g CO2

d)

amount excess = mass initially available - mass needed to react

start with mass of limiting reactant (18.5 g O2)

convert to grams of C2H2

use molar mass

18.5gO2(1molO2/32gO2)=0.578125 mol O2

use coefficients

0.578125molO2(2molC2H2/5molO2)=0.23125 mol C2H2

use molar mass

0.23125 mol C2H2(34.04gC2H2/1molC2H2)=7.87g C2H2 needed to react

amount excess = mass initially available - mass needed to react

amount excess = 13.7gC2H2 - 7.87 g C2H2

amount excess=5.83 g C2H2