zinc metal and hydrochloric acid react together according to the following equation 2HCl (aq)+ Zn(s) = ZnCl2(aq) +H2(g) if 5.98 g Zn reacts with excess HCl at 298 K and 0.978 atm what volume of H2 can be collected

2.29 L

Explanation:

In this question we have to start with the chemical reaction:

The reaction is already balanced. So, if we have an excess of HCl the compound that would limit the production of would be Zn. So, we have to follow a few steps:

1) Convert from grams to moles (Using the atomic mass of Zn 65.38 g/mol).

2) Convert from moles of Zn to moles of (Using the molar mass 1 mol = 1 mol Zn).

3) Convert from mol of to volume (Using the ideal gas equation PV=nRT).

First step:

Second step:

Third step:

We have to remember that R = 0.082 , so:

I hope it helps!

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