The reaction is,
H2S + I2 --------------> 2 HI +S
Molar weight of H2S = 34 g per mol
Molar weight of HI =128 g per mol
Molar weight of I2 =254 g per mol
Moles of H2S in 49.2 g = 49.2 /34 mol = 1.447 mol
So according to stoichiometry of the reaction, number of I2 mols needed
= 1.447 mol
The mass of I2 needed = 1.447 mol x 254 g
Answer:
a. 473mL.
b. 79.38kg
c. 24.47lb of fat
d. 42.5g of N.
Explanation:
a. A qt is equal to 946mL. 0.500qt are:
0.500qt * (946mL / 1qt) = 473mL
b. 1lb is equal to 0.4536kg, 175lb are:
175lb *(0.4536kg / 1lb) = 79.38kg
c. The fat in kg of the athlete is:
74kg * 15% = 11.1kg of fat. In pounds:
11.1kg * (1lb / 0.4536kg) = 24.47lb of fat
d. The mass of nitrogen in the fertilizer is:
10.0oz * 15% = 1.5oz of N
1 oz is equal to 28.35g. 1.5oz are:
1.5oz * (28.35g / 1oz) = 42.5g of N
Answer
to do homework or not
Explanation:
you can figure it out in one second
I think the particle starts to move slowly around because no heat or pressure is applied to it
Answer:
7.28 moles Ag°
Explanation:
Cu° + 2 AgNO₃ => Cu(NO₃)₂ + 2Ag°
Given 7.28 moles 7.28 moles
To determine limiting reactant, divide the mole values by the respective coefficient of balanced equation. The resulting smallest value is the limiting reactant. Note: this is a short cut method for determining limiting reactant only. Once the limiting reactant is determined one must use the given mole values of the limiting reactant to solve problem. That is ...
Limiting reactant determination:
Cu° + 2 AgNO₃ => Cu(NO₃)₂ + 2Ag°
Cu: 7.28 / 1 = 7.28
AgNO₃ : 7.28 / 2 = 3.64 => Limiting Reactant is AgNO₃
Solving Problem depends on AgNO₃; Cu will be in excess.
Since coefficients of AgNO₃ & Ag° are equal, then the moles AgNO₃ used equals moles Ag° produced and is therefore 7.28 moles Ag°.
