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Question: Explain how you will find the mass of sodium needed to make \( 5.68L \) of hydrogen gas STP after th...

Explain how you will find the mass of sodium needed to make 5.68L5.68L of hydrogen gas STP after the reaction, described in the following equation:
2Na+2H2O2NaOH+H22Na + 2H_2O \to 2NaOH + H_2

Explanation

Solution

We need to make sure that the given equation is always balanced, if not, we should balance it. Only from a balanced equation we can correctly associate the number of moles of the target elements, which are sodium and hydrogen here. The steps we follow to get the final answer would be to first find the total number of moles required from hydrogen’s volume, then to find mass of sodium by the molar ratio.

Complete step by step answer:
Let us look at what is given to us;
The volume of hydrogen gas at STP required =5.68L= 5.68L
To find, the mass of sodium =?= ?
We also have a balanced equation provided in the question, let us note it down again;
2Na+2H2O2NaOH+H22Na + 2H_2O \to 2NaOH + H_2
Now we need to understand from the balanced equation that, from the left hand side, 22 moles of sodium ( 2Na2Na ) is able to make 11 mole of hydrogen gas ( H2H_2 ) as given on the right hand side.
We need to know the relationship between mole and litres;
By result: 22.4L22.4L of volume comprises 11 mole of a gas
If that is the case then; 1L1L is equivalent to 1  mol22.4  L\dfrac{{1\;mol}}{{22.4\;L}}
Consequently, for 5.68L5.68L of hydrogen gas there will be 5.68  L22.4  L/mol\dfrac{{5.68\;L}}{{22.4\;L/mol}} moles of hydrogen.
5.68L  H2\therefore 5.68L\;H_2 contains 0.2540.254 moles of H2H_2 .
Since we know from the balanced equation that the ratio of Na:H2Na:H_2 is 2:12:1 , for sodium we need twice the number of moles, that is 0.5070.507 moles.
Let us consider sodium’s molecular weight now, that is 23  g/mol23\;g/mol
Therefore to find mass of sodium we need to apply a formula:
Mass  =Number  of  moles×Molecular  weightMass\; = Number\;of\;moles \times Molecular\;weight
Mass  of  Na=0.507mol×23g.mol1Mass\;of\;Na = 0.507mol \times 23g.mo{l^{ - 1}}
\therefore Mass  of  NaMass\;of\;Na needed to make 5.68L  of  H25.68L\;of\;H_2 =11.67g= 11.67g

Note: The mole concept is a widely used concept in Chemistry. Mole is similar to generic units like dozen, pair, etc. A mole gives an accurate quantity for the number of atoms or molecules in the given matter. A universally accepted constant, Avogadro’s number is said to be the number of entities inside a mole, the number has been calculated to be 6.023×10236.023 \times {10^{23}} . In a balanced equation the number given to the left of the components in the equation is said to describe the number of moles of that component required to produce the entities on the product side.