If 4.52 moles of na react with an excess of water, how many grams of h2 are formed?

4.52g

Explanation:

Firstly, to solve this problem, we need to write a balanced chemical reaction equation for sodium reacting with water

2Na+2H2O→2NaOH+H2

It can be seen that moles of sodium reacted with 2 moles of water to give 2 moles of sodium hydroxide and one mole of hydrogen gas. In this question, we are concerned with the sodium and hydrogen. Primarily, we can see that two moles of sodium yielded one mole of hydrogen gas. This means that the mole ratio of the sodium to the hydrogen is 2 to 1 or 2:1

This means we will require two moles of sodium for any one mole of hydrogen to be produced.

Hence, 4.52 moles of sodium means we will produce 4.52/2 moles of hydrogen which equals 2.26 moles.

Now we know that 2.26 moles of hydrogen is yielded. What we need do now is to get the mass of hydrogen gas produced. The mass produced is simply the number of moles of hydrogen multiplied by the molecular mass of the hydrogen gas. The molar mass of hydrogen gas is 1 + 1 which equals 2g/mol

Hence the mass of hydrogen gas produced is 2.26 * 2 = 4.52g

answer: - first one is the correct answer. .

explanations: - potassium sulfide is and cobalt(ii) chloride is .

charge on k is +1, charge on s is -2, charge on co is +2 and charge on cl is -1.

in a double replacement reaction the ion exchange takes place. in general the reaction looks like:

if one of the product is insoluble as per solubility rules then a precipitate of it is formed.

when we mix the aqueous solutions of potassium sulfide and cobalt(ii) chloride then chloride ion goes with potassium ion and sulfide ion goes with cobalt ion and the balanced equation is written as:

k and cl has equal charges, similarly co and s has equal charges and hence no subscripts are used for the product side.

as per the solubility rules, cos is insoluble and so it's precipitate is formed.

looking at the given choices, first choice is correct.