As you climb through the troposphere, the temperature:

6 NaOH + 2 Al --> 2 Na3AlO3 + 3 H2 If you start with 258 g of Al, how many moles of Na3AlO3 can you make?

tatyana61 [14]

Solution :

Assuming all Al will be consumed in the reaction.

From the given equation, 2 mole of Al produce 2 mole of $Na_3AlO_3$ .

So, 1 mole of Al produce 1 mole of $Na_3AlO_3$ . .....statement 1)

Number of moles of Al in 258 gram of Al.

$n = \dfrac{mass}{molar\ mass}\\\\n = \dfrac{258}{27}\ moles\\\\n = 9.556 \ moles$

Therefore, from statement 1) number of moles of $Na_3AlO_3$ produced is also 9.556 moles.

7 0

3 years ago

Which has more energy: yellow or violet light? And why?

Ilya [14]

This is funny, because last year I did a project on wave lengths, but despite that; Violet light has more energy than yellow light because on the light spectrum, violet has the least wavelengths. This means they carry the most energy.

8 0

3 years ago

Problem page gaseous ethane ch3ch3 will react with gaseous oxygen o2 to produce gaseous carbon dioxide co2 and gaseous water h2o

iVinArrow [24]

$m(\text{CO}_2) = 2.24 \; \text{g}$

Ethene react with oxygen at a $2 : 7$ molar ratio:

$2\; \text{C}_2 \text{H}_6 (g) + 7\; \text{O}_2 (g) \to 6\; \text{H}_2{O} (g) + 4\; \text{CO}_2 (g)$

Convert the quantity of each reactant supplied to number of moles of particles:

$n(\text{C}_2\text{H}_6) = 0.60 \; \text{g} / 28.05 \; \text{g} \cdot \text{mol}^{-1} = 0.0214 \; \text{mol}$
$n(\text{O}_2) = 3.27 \; \text{g} / 32.00 \; \text{g} \cdot \text{mol}^{-1} = 0.102 \; \text{mol}$

The question stated not whether both reactants were used up in this process. Thus start by testing the assumption that e.g., ethene was used up while some oxygen gas were left unreacted (ethene as the <em>limiting </em>reagent.) Under this assumption, the relative availability of the two species, $n(\text{C}_2 \text{H}_6) /2$ and $n(\text{O}_2) /7$ (as seen in the balanced chemical equation) shall satisfy the relationship

$n(\text{O}_2) / 7 - n(\text{C}_2 \text{H}_6) / 2 > 0$

In other words,

$n(\text{O}_2)/7 > n(\text{C}_2 \text{H}_6)/2$

$n(\text{C}_2 \text{H}_6) / n(\text{O}_2) < 2/7 \approx 0.286$

Evaluating the expression $n(\text{C}_2 \text{H}_6) / n(\text{O}_2)$ with data given in the question yields approximately $0.210 < 0.286$ , which does satisfy the relationship. Hence the assumption holds and ethene is the limiting reactant.

The quantity of a reactant produced in a chemical reaction is related to its stoichiometric (of relating to proportions) relationship with the limiting reactant (or any of the reactants in case of more than one limiting reactant.) For this scenario, given the molar ratio $n(\text{C}_2\text{H}_6) : n( \text{CO}_2) = 2:4$ ,