2Al + 6HCl ----> 2AlCl3 + 3H2
Here's how to do it:
<span>Balanced equation first: </span>
<span>Mg + HCl = H2 + MgCl2 unbalanced </span>
<span>Mg + 2 HCl = H2 = MgCl balanced </span>
<span>Therefore 1 mole Mg reacts with 2 moles Hcl. </span>
<span>50g Mg = ? moles (a bit over 2; you work it out) </span>
<span>75 g HCl = ? moles (also a bit over 2; you work it out) </span>
<span>BUT, you need twice the moles HCl; therefore it is the Mg that is in excess. (you can now work out how many moles are in excess, and therefore how much mg is left over). </span>
<span>So, 2 moles HCl produce 1 mole H2(g) </span>
<span>therefore, the amount of H2 produced is half the number of moles of HCl </span>
<span>At STP, there are X litres per mole of gas (look it up - about 22 from memory) </span>
<span>Therefore, knowing the moles of H2, you can calculate the volume</span>
Answer:
6.02×10²³ particles / 1 mol
Explanation:
In this question the choices are missed.
For determine the particles of Na, in 0.25 moles we should know that the mole referrs to the Avogadro's number.
1 mol of anything contains 6.02×10²³ entities, that's why we should apply this conversion factor to find the final answer:
0.25 mol . 6.02×10²³ particles / 1 mol = 1.50×10²³ particles
In conclussion: 1.50×10²³ particles of Na are contained in 0.25 moles of Na
