What volume of "laughing gas" (N2O) will be produced from 60 g of nitrogen gas (N2) and 85 g of oxygen gas (O2)? How much of the excess reactant will be left? 2N2 + O2 -> 2N2O

94.16g of N2O

50.88g of O2

Explanation:

O2 reacts with N2 as follows:

2N2 + O2 → 2N2O

To find moles of N2O produced we need to convert the mass of each gas to moles to find limiting reactant and the moles of N2O that could be produced:

Moles N2 (Molar mass: 28g/mol):

60g N2 * (1mol / 28g) = 2.14 moles N2

Moles O2 (Molar mass: 32g/mol):

85g O2 * (1mol / 32g) = 2.66 moles O2

For a complete reaction of 2.14 moles of N2 are needed:

2.14 mol N2 * (1mol O2 / 2 mol N2) = 1.07 moles of O2.

As there are 2.66 moles of O2, reactant in excess is O2 and will remain:

2.66 moles - 1.07 moles =

1.59 moles O2. The mass is:

1.59 moles O2 * (32g / mol) =

And there are produced:

2.14 mol N2 * (2 moles N2O / 2 moles N2) = 2.14 moles N2O.

The mass is (44g/mol for N2O):

2.14 moles N2O * (44g/mol) =

the answer would be a

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2