How many moles of nacl are present in 1.63 liters of a 0.954 m solution?

1.56 mol

Explanation:

c = n/V     Multiply both sides by V and transpose

n = Vc

c = 0.954 mol·L⁻¹

V =1.63 L    Calculate the moles

n = 0.954 × 1.63

n = 1.56 mol

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