Answer:
Explanation:
H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)
mole of NaOH = 23.6 * 10 ⁻³L * 0.2M
= 0.00472mole
let x be the no of mole of H3PO4 required of 0.00472mole of NaOH
3 mole of NaOH required ------- 1 mole of H3PO4
0.00472mole of NaOH ----------x
cross multiply
3x = 0.0472
x = 0.00157mole
[H3PO4] = mole of H3PO4 / Vol. of H3PO4
= 0.00157mole / (10*10⁻³l)
= 0.157M
<h3>The concentration of unknown phosphoric acid is 0.157M</h3>
1 mole --------- 22.4 ( at STP )
2.66 moles ---- ?
V = 2. 66 * 22.4 / 1
V = 59.584 / 1
V = 59.584 L
hope this helps!
Answer:
158.5g Zn are produced
Explanation:
To solve this question we have to find the moles of Aluminium. With the moles of Aluminium and the balanced reaction we can find the moles of Zn and its mass as follows:
<em>Moles Al -Molar mass: 26.98g/mol</em>
43.6g Al* (1mol/26.98g) = 1.616 moles Al
<em>Moles Zn:</em>
1.616 moles Al * (3mol Zn / 2mol Al) =
2.424 moles Zn are produced
<em>Mass Zn -Molar mass: 65.38g/mol-</em>
2.424 moles Zn * (65.38g / mol) =
<h3>158.5g Zn are produced</h3>