2.5 mole O2 should react (2 * 2.5) = 5.0 mole H2 but there is 6.0 mole H2. H2 remain in excess. O2 is the limiting reactant. mole of H2O formed = 2 * 2.5 = 5.0 mole.

5 0

3 years ago

Copper(II) sulfide is formed when copper and sulfur are heated together. In this reaction,

storchak [24]

Answer:

The mass of copper(II) sulfide formed is:

= 81.24 g

Explanation:

The Balanced chemical equation for this reaction is :

$Cu(s) + S\rightarrow CuS$

given mass= 54 g

Molar mass of Cu = 63.55 g/mol

$Moles = \frac{given\ mass}{Molar\ mass}$

$moles=\frac{54}{63.55}$

Moles of Cu = 0.8497 mol

Given mass = 42 g

Molar mass of S = 32.06 g/mol

$Moles = \frac{given\ mass}{Molar\ mass}$

$moles=\frac{42}{32.06}$

Moles of S = 1.31 mol

Limiting Reagent : The reagent which is present in less amount and consumed in a reaction

First find the limiting reagent :

$Cu + S\rightarrow CuS$

1 mol of Cu require = 1 mol of S

0.8497 mol of Cu should require = 1 x 0.8497 mol

= 0.8497 mol of S

S present in the reaction Medium = 1.31 mol

S Required = 0.8497 mol

S is present in excess and Cu is limiting reagent

All Cu is consumed in the reaction

Amount Cu will decide the amount of CuS formed

$Cu + S\rightarrow CuS$

1 mole of Cu gives = 1 mole of Copper sulfide

0.8497 mol of Cu = 1 x 0.8497 mole of Copper sulfide

= 0.8497

Molar mass of CuS = 95.611 g/mol

$Moles = \frac{given\ mass}{Molar\ mass}$

$0.8497 = \frac{given\ mass}{95.611}$

Mass of CuS = 0.8497 x 95.611

= 81.24 g

3 0

3 years ago