Question

Calcium nitrate and ammonium flouride react to form calcium flouride, dinitrogen monoxide, and water vapor. What mass of each substance is present after 21.83g of calcium nitrate and 22.75g of ammonium flouride react completely?

when 75.5g of calcium and 43.1g of nitrogen gas undergo a reaction that has a 88% yield, what mass of calcium nitride forms?

when 73.8g of calcium and 41.8g of nitrogen gas undergo a reaction that has a 92.4% yield, what mass of calcium nitride forms?

a sample of 0.767 mol of a metal m reacts completely with excess flourine to form 59.9g of MF2. what element is represented by the symbol M?

when 29.8g of methane and 47.6 g of chlorine gas undergo a reaction that has a 74.5% yield, what mass of chlormethane (CH3cl) forms?

Answer 1

Ca(NO₃)₂ + 2 NH₄F → CaF₂ + 2 N₂O + 4 H₂O

Molecular weight of Ca(NO₃)₂ = 164.088 g/mol

Molecular weight of NH₄F = 37.037 g/mol

Molecular weight of CaF₂ = 78.07 g/mol

Molecular weight of N₂O = 44.013 g/mol

Molecular weight of H₂O = 18 g/mol.

a) What mass of each substance is present after 21.83g of calcium nitrate and 22.75g of ammonium flouride react completely?

Moles of Calcium nitrate = 21.83g/ 164.088gmo^-1 = 0.1330 mole

Moles of Ammonium flouride = 22.75g/ 37.037gmol^-1 = 0.614 mole

One mole of Calcium nitrate reacts with two moles of Ammonium fluoride

so, 0.1330 moles of calcium nitrate reacts with 0.2660 moles ammonium fluoride.

Limiting reagent is calcium nitrate and excess reagent is ammonium fluoride.

0.1330 moles of calcium nitrate reacts with 0.2660 moles ammonium fluoride reacts to give 0.1330 moles of CaF₂, 0.2660 moles of N₂O and 0.532 moles of water.

CaF₂ = 0.1330 mole x 78.07 g/mole = 10.38 g

N₂O = 0.2660 mole x 44.013 g/mole = 11.70 g

CaF₂ = 0.532 mole x 18 g/mole = 9.576 g

b) when 75.5g of calcium and 43.1g of nitrogen gas undergo a reaction that has a 88% yield, what mass of calcium nitride forms?

3Ca + N₂ → Ca₃N₂

moles of Ca = 75.5g/40g mol^-1 = 1.887 mol

moles of N₂ = 43.1g/28gmol^-1 = 1.539 mol

so 1.887 mol/3 (0.629 mol) of Calcium reacts with 0.629 mole of N₂ to give 0.629 moles of calcium nitride i.e. 0.629 moles x 148.25g/mol = 93.24 g

Yield of 80% corresponds to 0.80 * 93.24 g = 74.6 g

c)when 73.8g of calcium and 41.8g of nitrogen gas undergo a reaction that has a 92.4% yield, what mass of calcium nitride forms?

3Ca + N₂ → Ca₃N₂

moles of Ca = 73.8g/40g mol^-1 = 1.845 mol

moles of N₂ = 41.8g/28gmol^-1 = 1.492 mol

so 1.845 mol/3 (0.615 mol) of Calcium reacts with 0.615 mole of N₂ to give 0.615 moles of calcium nitride ie 0.615 moles x 148.25g/mol = 91.17 g

Yield of 92.4% corresponds to 0.924 * 91.17 g = 84.24 g

d) a sample of 0.767 mol of a metal M reacts completely with excess flourine to form 59.9g of MF2. what element is represented by the symbol M?

M + F₂---------> MF₂

The reaction 1:1 in mole ratio, so 0.737 moles of Metal reacts with 0.737 moles of F₂ to 0.737 moles of MF₂.

0.737mole corresponds to 59.98 g, so one mole is (59.98g/0.737mol) 78 g/mol

So the atomic weight of M is (78g/mol - 38g/mol) = 40g , so the element is Ca and the compound is CaF₂

e) when 29.8g of methane and 47.6 g of chlorine gas undergo a reaction that has a 74.5% yield, what mass of chlormethane (CH3cl) forms?

CH₄ + Cl₂ ------------------> CH₃Cl + HCl

mole of methane = 29.8g/ 16gmol^-1 =1.862 mol

mole of chlorine = 47.6g/ 71gmol^-1 =0.670 mol

Amount of CHCl₃ produced = 0.670 moles = 0.670 moles x50.5 g/mole = 33.83 g

Yield of 74.5% corresponds to 0.745 * 33.83 g = 25.20 g of CH₃Cl