Question

5. Calculate the mass of sodium nitrate, required to produce 128 g of oxygen. 6. What is the limiting reagent when 0.50 mol of Cr and 1.0 mol of H3PO4 react according to the following chemical equation? 2Cr + 2H3PO4 → 2CrPO4 + 3H2 If 0.20 mol of CrPO4 is recovered from the reaction described above, what is the percent yield? 7. What volume of 0.0125 M hydrogen bromide solution is required to titrate 250 ml of a 0.0100 M calcium hydroxide solution?

Answer 1

5.

We need a stoichiometric equation:

$NaNO_3(s) + energy~NaNO_3+1/2 O_2$

We require a 128.g

 mass of dioxygen gas, i.e. 

$128.g/32.00.g=4.g$ /mol

Given the equation, clearly we need AT LEAST 8 mol of sodium nitrate i.e

$8mol×85g/mol$

1 mol= 680.grams.

6.Limiting reagent is reagent that is in a smaller molar amount in a chemical reaction. Limiting reagent limits the amount of products that can be made. In the reaction shown above the limiting reagent is Cr (0.50 mole < 1.0 mole).

$percent \space yield = \frac {actual \space yield }{theoretical \space yield} \cdot 100 \%$

$actual \space yield = 0.20 \space mol \cdot \frac {146.97 g/mol} {1mol} = 29 g$

Calculate theoretical yield:

$mol \space of \space CrPO_4 = 0.50 \space mol \space Cr \cdot \frac {2mol \space CrPO_4}{2 mol \space Cr}=0.50 mol$

$theoretical \space yield = mass \space of \space CrPO_4 = 0.50 \cdot 146.97 \space g/mol = 73 \space g$

$percent \space yield = \frac {29 g}{73g} \cdot 100 \% =40 \%$

7.$V=\dfrac{0.0100 mol/L*250 ml}{0.0125mol/L}=200mL$