Answer:
1.19 g
Explanation:
Given that:
Molecular weight of diene 136 g/mol
Molecular weight of Maleic Anhydride 96 g/mol
Mass of crude diene = 2.5 g
Percent of Composition for Peak A = 66%
Percent of Composition for Peak B = 22.66%
Percent of Composition for Peak C = 11.33%
Let us determine the amount of diene in the sample;
So, using to represent the amount of diene; we have :
=
=
= 1.65 g
Using the limiting reagent equation to determine the amount of Maleic Anhydride
=
=
= 1.19 g
The grams of Maleic Anhydride used to react with 2.5 g sample of crude diene = 1.19 g
Answer:
(3) Equals the mass of CaO plus the mass of CO₂
Explanation:
The important principle to remember is the Law of Conservation of Mass — the total mass of the reactants must equal the total mass of the products.
Reactants ⟶ products
CaCO₃ ⟶ CaO + CO₂
Mass of CaCO₃ = Mass of CaO + mass of CO₂
Answer:
99.56 g.
Explanation:
- The balanced reaction is:
<em>3Ca(NO₃)₂ + 2Na₃PO₄ → 6NaNO₃ + Ca₃(PO₄)₂.</em>
It is clear that 3.0 moles of Ca(NO₃)₂ react with 2.0 moles of Na₃PO₄ to produce 6.0 moles of NaNO₃ and 1.0 mole of Ca₃(PO₄)₂.
- We need to calculate the no. of moles of 96.1 g of Ca(NO₃)₂:
n = mass/molar mass = (96.1 g)/(164.088 g/mol) = 0.5857 mol.
<u><em>Using cross multiplication:</em></u>
3.0 moles of Ca(NO₃)₂ produce → 6.0 moles of NaNO₃.
0.5857 moles of Ca(NO₃)₂ produce → ??? moles of NaNO₃.
∴ The no. of moles of NaNO₃ = (6.0 mol)(0.5857 mol)/(3.0 moles) = 1.171 mol.
<em>∴ The no. of grams of NaNO₃ = (no. of moles of NaNO₃)(molar mass)</em> = (1.171 mol)(84.99 g/mol) = <em>99.56 g.</em>
Answer:
B. When it is obtained by rounding, as in 3.0
Explanation: