Answer:
8.0356 * 10^-5 moles of NaHCO3
Explanation:
Sulphuric acid = H2SO4
Sodium bicarbonate = NaHCO3
The reaction between both compounds is given by;
2NaHCO3(aq) + H2SO4(aq) → Na2SO4(aq) + 2CO2(g) + 2H2O(l)
In the reactin above;
2 mol of NaHCO3 neutralizes 1 mol of H2SO4
At stp, 1 mol occupies 22.4 L;
1 mol = 22.4 L = 22400 mL
x mol = 0.9 mL
x = 0.9 / 22400 = 4.0178 * 10^-5 moles of H2SO4
Since 2 mol = 1 mol from the equation;
x mol = 4.0178 * 10^-5
x mol = 2 * 4.0178 * 10^-5
x = 8.0356 * 10^-5 moles of NaHCO3
Answer: ABC&AMN are congruent by ASA comgruence
angle A=A
sideAN=AB
angle M=C
Explanation:
Answer:
True
Explanation:
Its used to measure mass. Ten gallons weigh 8 pounds.
:D
Have a good day flopper :D
<em></em>
<em>*eliza*</em>
4.20 mol Al would react completely with 4.20 x (1/2) = 2.10 mol Fe2O3, but there is not that much Fe2O3 present, so Fe2O3 is the limiting reactant. (1.75 mol Fe2O3) x (2/1) x ( 55.8452 g Fe/mol) = 195 g Fe 3 MgO + 2 H3PO4 → Mg3(PO4)2 + 3 H2O (15.0 g MgO) / (40.3045 g MgO/mol) = 0.37217 mol MgO (18.5 g H3PO4) / (97.9953 g H3PO4/mol) = 0.18878 mol H3PO4 0.18878 mol H3PO4 would react completely with 0.18878 x (3/2) = 0.28317 mole of MgO, but there is more MgO present than that, so MgO is in excess and H3PO4 is the limiting reactant. Now we must consider why the problem tells us "17.6g of Mg3(PO4)2 is obtained". The first possibility is that it's just there for the sake of confusion -- in which case ignore it and proceed this way: ((0.37217 mol MgO initially) - (0.28317 mole MgO reacted)) x (40.3045 g MgO/mol) = 3.59 g MgO left over However, if the amount of magnesium phosphate obtained is given because the reaction was stopped before it was complete, the amount obtained governs the amount reacted and the amount left over, so proceed this way: (17.6g Mg3(PO4)2) / (262.8581 g Mg3(PO4)2/mol) x (3/1) = 0.20087 mol MgO reacted ((0.37217 mol MgO initially) - (0.20087 mole MgO reacted)) x (40.3045 g MgO/mol) = 6.90 MgO left over
