Balanced equation first.
2H2 + O2 -> 2H2O
250 moles O2 (2 mole H2O/1 mole O2)
= 500 mole H2O produced
now, since I am forgetful, I will use density formula at 25 C
Density = grams/milliliters
500 moles H2O (18.016 grams/1 mole H2O)
= 9008 grams H2O
0.9982 g/ml = 9008 grams/milliliters
9024.24 milliliters H2O
this is......
9.02 liters of water produced in this reaction.
90g water
24
If everything is under gaseous condition and with equal pressure and temperature, then also 4.0 L SO3 gas is produced from 4.0 L SO2,g.
117(g K) * [1.008(g/mol H) / 39.098(g/mol K)] = 3.02 gram hydrogen
I assume you mean this reaction. Mg + 2HCl -> MgCl2 + H2 need moles HCl as it limits Molarity = Moles of solute/Liters of solution ( 250.0 ml = 0.250 L ) 3. 0 M HCl = moles HCl/0.250 liters = 0.75 moles HCl ----------------------------now, refer to equation and drive reactant against product to get moles H2 0.75 moles HCl (1 mole H2/2 mole HCl) = 0.375, or 0.38 mole hydrogen gas produced ------------------------------------------------------------
11.45
24
60 grams.
If everything is under gaseous condition and with equal pressure and temperature, then also 4.0 L SO3 gas is produced from 4.0 L SO2,g.
2 volumes of ammonia gas.
24 real answer,bra
Assuming we are at standard temperature and pressure (STP), the answer is approximately 1.85L of hydrogen gas. The HCl is the limiting reactant and the Mg is the excess reactant.
117(g K) * [1.008(g/mol H) / 39.098(g/mol K)] = 3.02 gram hydrogen
I assume you mean this reaction. Mg + 2HCl -> MgCl2 + H2 need moles HCl as it limits Molarity = Moles of solute/Liters of solution ( 250.0 ml = 0.250 L ) 3. 0 M HCl = moles HCl/0.250 liters = 0.75 moles HCl ----------------------------now, refer to equation and drive reactant against product to get moles H2 0.75 moles HCl (1 mole H2/2 mole HCl) = 0.375, or 0.38 mole hydrogen gas produced ------------------------------------------------------------
depends
11.45
the ratio between HCL and CL2 is 4:2 V(Cl2)=0.98/2=0.49L
2/1